$3^p-2^p$ squarefree? The Next CEO of Stack OverflowIs $(3^p-1)/2$ always squarefree?Square-free integer$fracq^p-1q-1$ squarefree?How to prove there are an infinite number of squarefree numbers of the form $2^p-1$?Are [Wieferich] primes the only solutions to $2^n-1 equiv 1 pmodn^2$?Is the product of two primes ALWAYS a semiprime?Is there a conjecture with maximal prime gapsCounting squarefree 3-almost primes. (Solved!)Why is every prime number ($5$ and higher) divisible by $24$ (into a whole integer) when you square it and subtract $1$?Alternative proof of infinitude of prime.What is the Largest Possible Prime Number? sic The largest possible prime set?For which natural numbers are $phi(n)=2$?The hunting of “missing primes”
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$3^p-2^p$ squarefree?
The Next CEO of Stack OverflowIs $(3^p-1)/2$ always squarefree?Square-free integer$fracq^p-1q-1$ squarefree?How to prove there are an infinite number of squarefree numbers of the form $2^p-1$?Are [Wieferich] primes the only solutions to $2^n-1 equiv 1 pmodn^2$?Is the product of two primes ALWAYS a semiprime?Is there a conjecture with maximal prime gapsCounting squarefree 3-almost primes. (Solved!)Why is every prime number ($5$ and higher) divisible by $24$ (into a whole integer) when you square it and subtract $1$?Alternative proof of infinitude of prime.What is the Largest Possible Prime Number? sic The largest possible prime set?For which natural numbers are $phi(n)=2$?The hunting of “missing primes”
$begingroup$
Original question:
$3^n-2^n-1$ seems to be squarefree. Is it?
Answer: No, but amongst the primes dividing one of these numbers, $23$ seems to be a special case: no $23^2$ divide any of them
Are there other case where $p$ divide a $3^n-2^n-1$ but $p^2$ does not?
Edited question, based on answers:
It is conjectured that $2^p-1$ is squarefree.
Could it be that $3^p-1$ is also squarefree for $pneq2$ and $5$ (where $11^2$ appears)?
Could it be that $3^p-2^p$ is also squarefree for $pneq11$ (where $23^2$ appears)?
Thanks
edit: just saw this Is $(3^p-1)/2$ always squarefree?.
Note: $n$ is a positive natural natural in the original question and $p$ is a prime in my edit.
number-theory prime-numbers prime-factorization
asked Mar 19 at 20:40
Collag3nCollag3n
784211
$endgroup$
|
show 2 more comments
$begingroup$
Original question:
$3^n-2^n-1$ seems to be squarefree. Is it?
Answer: No, but amongst the primes dividing one of these numbers, $23$ seems to be a special case: no $23^2$ divide any of them
Are there other case where $p$ divide a $3^n-2^n-1$ but $p^2$ does not?
Edited question, based on answers:
It is conjectured that $2^p-1$ is squarefree.
Could it be that $3^p-1$ is also squarefree for $pneq2$ and $5$ (where $11^2$ appears)?
Could it be that $3^p-2^p$ is also squarefree for $pneq11$ (where $23^2$ appears)?
Thanks
edit: just saw this Is $(3^p-1)/2$ always squarefree?.
Note: $n$ is a positive natural natural in the original question and $p$ is a prime in my edit.
number-theory prime-numbers prime-factorization
asked Mar 19 at 20:40
Collag3nCollag3n
784211
$endgroup$
1
$begingroup$
wolframalpha.com/input/… Nope, but counterexamples seem to be pretty sparse.
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– user113102
Mar 19 at 20:49
1
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Use computer simulations to have an idea about it...
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– Jean Marie
Mar 19 at 20:51
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Thx user113102, I sometimes used wolfram to plot things but didn't thought about using it that way.
$endgroup$
– Collag3n
Mar 19 at 20:54
$begingroup$
Deleted and reposted because I cannot freely edit my OWN comments: For the original problem, $23$ is not unique. There are lots of primes p>3 for which p2 never divides $3^n−2n−1$, the smallest of these actually being $5$. Just to be clear, in the edited question the exponent is now prime, right?
$endgroup$
– Oscar Lanzi
Mar 20 at 22:56
$begingroup$
Yes, sorry, I edited the post. Like Will, I was only looking at prime dividing these numbers, and indeed, I only consider prime exponent in the edited question (since there are clearly squares in the 3 forms for other $n$)
$endgroup$
– Collag3n
Mar 21 at 7:10
|
show 2 more comments
$begingroup$
Original question:
$3^n-2^n-1$ seems to be squarefree. Is it?
Answer: No, but amongst the primes dividing one of these numbers, $23$ seems to be a special case: no $23^2$ divide any of them
Are there other case where $p$ divide a $3^n-2^n-1$ but $p^2$ does not?
Edited question, based on answers:
It is conjectured that $2^p-1$ is squarefree.
Could it be that $3^p-1$ is also squarefree for $pneq2$ and $5$ (where $11^2$ appears)?
Could it be that $3^p-2^p$ is also squarefree for $pneq11$ (where $23^2$ appears)?
Thanks
edit: just saw this Is $(3^p-1)/2$ always squarefree?.
Note: $n$ is a positive natural natural in the original question and $p$ is a prime in my edit.
number-theory prime-numbers prime-factorization
asked Mar 19 at 20:40
Collag3nCollag3n
784211
$endgroup$
Original question:
$3^n-2^n-1$ seems to be squarefree. Is it?
Answer: No, but amongst the primes dividing one of these numbers, $23$ seems to be a special case: no $23^2$ divide any of them
Are there other case where $p$ divide a $3^n-2^n-1$ but $p^2$ does not?
Edited question, based on answers:
It is conjectured that $2^p-1$ is squarefree.
Could it be that $3^p-1$ is also squarefree for $pneq2$ and $5$ (where $11^2$ appears)?
Could it be that $3^p-2^p$ is also squarefree for $pneq11$ (where $23^2$ appears)?
Thanks
edit: just saw this Is $(3^p-1)/2$ always squarefree?.
Note: $n$ is a positive natural natural in the original question and $p$ is a prime in my edit.
number-theory prime-numbers prime-factorization
number-theory prime-numbers prime-factorization
asked Mar 19 at 20:40
Collag3nCollag3n
784211
asked Mar 19 at 20:40
Collag3nCollag3n
784211
edited Mar 21 at 7:04
Collag3n
asked Mar 19 at 20:40
Collag3nCollag3n
784211
asked Mar 19 at 20:40
Collag3nCollag3n
784211
asked Mar 19 at 20:40
Collag3nCollag3n
784211
784211
1
$begingroup$
wolframalpha.com/input/… Nope, but counterexamples seem to be pretty sparse.
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– user113102
Mar 19 at 20:49
1
$begingroup$
Use computer simulations to have an idea about it...
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– Jean Marie
Mar 19 at 20:51
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Thx user113102, I sometimes used wolfram to plot things but didn't thought about using it that way.
$endgroup$
– Collag3n
Mar 19 at 20:54
$begingroup$
Deleted and reposted because I cannot freely edit my OWN comments: For the original problem, $23$ is not unique. There are lots of primes p>3 for which p2 never divides $3^n−2n−1$, the smallest of these actually being $5$. Just to be clear, in the edited question the exponent is now prime, right?
$endgroup$
– Oscar Lanzi
Mar 20 at 22:56
$begingroup$
Yes, sorry, I edited the post. Like Will, I was only looking at prime dividing these numbers, and indeed, I only consider prime exponent in the edited question (since there are clearly squares in the 3 forms for other $n$)
$endgroup$
– Collag3n
Mar 21 at 7:10
|
show 2 more comments
1
$begingroup$
wolframalpha.com/input/… Nope, but counterexamples seem to be pretty sparse.
$endgroup$
– user113102
Mar 19 at 20:49
1
$begingroup$
Use computer simulations to have an idea about it...
$endgroup$
– Jean Marie
Mar 19 at 20:51
$begingroup$
Thx user113102, I sometimes used wolfram to plot things but didn't thought about using it that way.
$endgroup$
– Collag3n
Mar 19 at 20:54
$begingroup$
Deleted and reposted because I cannot freely edit my OWN comments: For the original problem, $23$ is not unique. There are lots of primes p>3 for which p2 never divides $3^n−2n−1$, the smallest of these actually being $5$. Just to be clear, in the edited question the exponent is now prime, right?
$endgroup$
– Oscar Lanzi
Mar 20 at 22:56
$begingroup$
Yes, sorry, I edited the post. Like Will, I was only looking at prime dividing these numbers, and indeed, I only consider prime exponent in the edited question (since there are clearly squares in the 3 forms for other $n$)
$endgroup$
– Collag3n
Mar 21 at 7:10
1
1
$begingroup$
wolframalpha.com/input/… Nope, but counterexamples seem to be pretty sparse.
$endgroup$
– user113102
Mar 19 at 20:49
$begingroup$
wolframalpha.com/input/… Nope, but counterexamples seem to be pretty sparse.
$endgroup$
– user113102
Mar 19 at 20:49
1
1
$begingroup$
Use computer simulations to have an idea about it...
$endgroup$
– Jean Marie
Mar 19 at 20:51
$begingroup$
Use computer simulations to have an idea about it...
$endgroup$
– Jean Marie
Mar 19 at 20:51
$begingroup$
Thx user113102, I sometimes used wolfram to plot things but didn't thought about using it that way.
$endgroup$
– Collag3n
Mar 19 at 20:54
$begingroup$
Thx user113102, I sometimes used wolfram to plot things but didn't thought about using it that way.
$endgroup$
– Collag3n
Mar 19 at 20:54
$begingroup$
Deleted and reposted because I cannot freely edit my OWN comments: For the original problem, $23$ is not unique. There are lots of primes p>3 for which p2 never divides $3^n−2n−1$, the smallest of these actually being $5$. Just to be clear, in the edited question the exponent is now prime, right?
$endgroup$
– Oscar Lanzi
Mar 20 at 22:56
$begingroup$
Deleted and reposted because I cannot freely edit my OWN comments: For the original problem, $23$ is not unique. There are lots of primes p>3 for which p2 never divides $3^n−2n−1$, the smallest of these actually being $5$. Just to be clear, in the edited question the exponent is now prime, right?
$endgroup$
– Oscar Lanzi
Mar 20 at 22:56
$begingroup$
Yes, sorry, I edited the post. Like Will, I was only looking at prime dividing these numbers, and indeed, I only consider prime exponent in the edited question (since there are clearly squares in the 3 forms for other $n$)
$endgroup$
– Collag3n
Mar 21 at 7:10
$begingroup$
Yes, sorry, I edited the post. Like Will, I was only looking at prime dividing these numbers, and indeed, I only consider prime exponent in the edited question (since there are clearly squares in the 3 forms for other $n$)
$endgroup$
– Collag3n
Mar 21 at 7:10
|
show 2 more comments
3 Answers
3
active
oldest
votes
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We can show that there are multiples of $49$ using elementary modular arithmetic techniques.
Let $p$ be a prime greater than $3$ (why?), and seek values of $n$ for which
$$3^nequiv 2^n-1bmod p^2$$
Multiply by $(3^-1)^n-1$ getting
$$3equiv (2×3^-1)^n-1bmod p^2$$
Let us first try $p=5$. We die because the left side is a nonquadratic residue $bmod p=5$ and the right side, with $2×3^-1equiv 4bmod 5$, is a quadratic residue.
Fortunately, for $p=7$ we avoid this contradiction because $2×3^-1$ is nonquadratic $bmod 7$ thus also nonquadratic $bmod 49$. We then have
$$3equiv 17^n-1bmod 49$$
where $17equiv 3bmod 7$ is a primitive root in the group of units $bmod 49$ (the only nonprimitive root $bmod 49$ congruent to $3bmod 7$ is $31$), therefore this equation must have positive whole number solutions for $n$.
Will Jagy has identified the minimal solution as $n=38$, so let us check this case $bmod 49$. Since units give $1$ when raised to the power of $42$, we may render
$$3^38equiv (3^-1)^4equiv 33^4equiv 11^2equiv 121equivcolorblue23bmod 49$$
And
$$2^37equiv (2^-1)^5equiv 25^5equiv 25×(-12)^2equiv 3600equiv colorblue23bmod 49$$
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
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Oscar, I realized I had a quick way to find squares of small primes, as I write my own factoring-related routines to get quick factors if needed: 712 +++ = 271^2 cdot mboxBIG 1588 +++ = 73 191^2 cdot mboxBIG 2340 +++ = 127^2 cdot mboxBIG 2531 +++ = 1021^2 cdot mboxBIG
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– Will Jagy
Mar 20 at 15:46
add a comment |
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Well, I posted too quickly.....for $n=67$ it seems we have 2 factor $11$
answered Mar 19 at 20:49
Collag3nCollag3n
784211
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4
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Also, $7^2 mid 3^38 - 2^37$.
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– Misha Lavrov
Mar 19 at 20:50
1
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Yes, That's because I missed that one that i posted....too quickly
$endgroup$
– Collag3n
Mar 20 at 8:38
add a comment |
$begingroup$
At the very end, I put a long list of these. The great majority of squared primes are below $100,$ but I did catch a few larger, so far $127, 191, 271, 1021 : ;$
712 +++ = 271^2 cdot mboxBIG
1588 +++ = 73 191^2 cdot mboxBIG
2340 +++ = 127^2 cdot mboxBIG
2531 +++ = 1021^2 cdot mboxBIG
There is a way to deal with this without doing genuine factoring. For example, your number is divisible by $49 = 7^2$ whenever $n = 42 k + 38.$
2 7 prime 7
8 6433 prime 7
14 4774777 prime 7
20 3486260113 prime 7
38 1350851580234038617 prime squared 49
80 147808829414345318853173402891795944513 prime squared 49
122 16173092699229880893715960009594875525837473033720099268457 prime squared 49
Your number is divisible by $121 = 11^2$ whenever $n = 110 k + 67.$
7 2123 prime 11
17 129074627 prime 11
27 7625530376123 prime 11
67 92709463147824050109467087204123 prime squared 121
177 2821383260958014531084804730393073172748132970923952481977527762896658545213494562627 prime squared 121
287 85861822891966292897565943276292392423908891501494514127947492094325821620603901184289283467528859878643948705742648123768724313989998123 prime squared 121
Your number is divisible by $289 = 17^2$ whenever $n = 272 k + 214.$
6 697 prime 17
22 31378962457 prime 17
38 1350851580234038617 prime 17
214 1270423474759653869629541561076150845942627861345583970679777076713806668073848513969400550694997546777 prime squared 289
486 7602033756829688179535612101927342434798006222913345882096671718462026450847558385638299237091029898106108915679642725019874527596206183615695170393375601813754958083630064304945006176793537681250116409274581709738622832815408017497 prime squared 289
758 45489491014727452017657094699993344217699387580459634625293727208970468768085650812024804178818092172750636489054819257623093543090018879398677204096366573883113560178809969964755425161715086488029060167818542608680433377185439106005853203184618487801892144052367301608312676367936439310746339829776474186019806821915640418802179770139744725661919759420980369817 prime squared 289
Surprising, your number is often divisible by $23$ but never by $529 = 23^2.$ Go Figure.
3 23 prime 23
14 4774777 prime 23
25 847271832227 prime 23
36 150094600937260753 prime 23
47 26588814288588759110123 prime 23
58 4710128697102129646845747817 prime 23
NO 23 SQUARED
Your number is divisible by $961 = 31^2$ whenever $n = 930 k + 828.$
18 387289417 prime 31
48 79766442936135021508033 prime 31
78 16423203268260507030504015972062417017 prime 31
828 prime squared 961
1758 prime squared 961
2688 prime squared 961
3618 prime squared 961
4548 prime squared 961
Your number is divisible by $1369 = 37^2$ whenever $n = 1332 k + 383.$
23 94138984523 prime 37
59 14130386091450504128613099323 prime 37
95 2120895147045314099684568958946760345244084523 prime 37
383 prime squared 1369
1715 prime squared 1369
3047 prime squared 1369
4379 prime squared 1369
5711 prime squared 1369
I also did a bounded factoring: given one of these numbers, use trial division with primes $p < 1200.$ I did catch a $1021^2$ this way..
jagy@phobeusjunior:~$ ./mse | grep "^"
38 = 7^2 17 cdot mboxBIG
67 = 11^2 cdot mboxBIG
80 = 7^2 23 607 cdot mboxBIG
122 = 7^2 137 599 cdot mboxBIG
164 = 7^3 113 cdot mboxBIG
177 = 11^2 cdot mboxBIG
206 = 7^2 41 cdot mboxBIG
214 = 17^2 cdot mboxBIG
248 = 7^2 cdot mboxBIG
287 = 11^2 cdot mboxBIG
290 = 7^2 47 809 1033 cdot mboxBIG
332 = 7^2 1193 cdot mboxBIG
374 = 7^2 17 1087 cdot mboxBIG
383 = 37^2 cdot mboxBIG
397 = 11^3 cdot mboxBIG
416 = 7^2 233 cdot mboxBIG
458 = 7^3 439 cdot mboxBIG
486 = 17^2 41 cdot mboxBIG
500 = 7^2 113 cdot mboxBIG
507 = 11^2 83 cdot mboxBIG
508 = 73^2 cdot mboxBIG
542 = 7^2 23 cdot mboxBIG
584 = 7^2 431 cdot mboxBIG
606 = 41^2 cdot mboxBIG
617 = 11^2 cdot mboxBIG
626 = 7^2 cdot mboxBIG
668 = 7^2 cdot mboxBIG
710 = 7^2 17 911 cdot mboxBIG
712 = 271^2 cdot mboxBIG
727 = 11^2 47 cdot mboxBIG
752 = 7^3 89 cdot mboxBIG
758 = 7 17^3 cdot mboxBIG
794 = 7^2 cdot mboxBIG
828 = 23 31^2 127 191 cdot mboxBIG
836 = 7^2 113 cdot mboxBIG
837 = 11^2 683 cdot mboxBIG
878 = 7^2 cdot mboxBIG
920 = 7^2 cdot mboxBIG
947 = 11^2 983 cdot mboxBIG
957 = 11 47^2 229 cdot mboxBIG
962 = 7^2 cdot mboxBIG
1004 = 7^2 23 937 cdot mboxBIG
1030 = 17^2 151 cdot mboxBIG
1046 = 7^3 17 41 cdot mboxBIG
1057 = 11^2 59 431 cdot mboxBIG
1088 = 7^2 cdot mboxBIG
1130 = 7^2 cdot mboxBIG
1167 = 11^2 cdot mboxBIG
1172 = 7^2 113 cdot mboxBIG
1214 = 7^2 569 cdot mboxBIG
1256 = 7^2 47 cdot mboxBIG
1277 = 11^2 cdot mboxBIG
1298 = 7^2 cdot mboxBIG
1302 = 17^2 47 223 263 cdot mboxBIG
1340 = 7^3 cdot mboxBIG
1382 = 7^2 17 cdot mboxBIG
1387 = 11^2 cdot mboxBIG
1424 = 7^2 479 cdot mboxBIG
1466 = 7^2 23 cdot mboxBIG
1491 = 83^2 157 cdot mboxBIG
1497 = 11^2 433 cdot mboxBIG
1508 = 7^2 113 cdot mboxBIG
1550 = 7^2 727 cdot mboxBIG
1574 = 7 17^2 cdot mboxBIG
1588 = 73 191^2 cdot mboxBIG
1592 = 7^2 cdot mboxBIG
1607 = 11^3 37 167 cdot mboxBIG
1634 = 7^5 cdot mboxBIG
1676 = 7^2 cdot mboxBIG
1715 = 37^2 587 cdot mboxBIG
1717 = 11^2 1117 cdot mboxBIG
1718 = 7^2 17 cdot mboxBIG
1758 = 31^2 cdot mboxBIG
1760 = 7^2 cdot mboxBIG
1802 = 7^2 cdot mboxBIG
1827 = 11^2 cdot mboxBIG
1844 = 7^2 113 919 cdot mboxBIG
1846 = 17^2 41 cdot mboxBIG
1886 = 7^2 41 863 cdot mboxBIG
1928 = 7^3 23 cdot mboxBIG
1937 = 11^2 cdot mboxBIG
1970 = 7^2 cdot mboxBIG
2012 = 7^2 cdot mboxBIG
2038 = 17 23 47^2 cdot mboxBIG
2047 = 11^2 cdot mboxBIG
2054 = 7^2 17 cdot mboxBIG
2096 = 7^2 cdot mboxBIG
2118 = 17^2 31 cdot mboxBIG
2138 = 7^2 cdot mboxBIG
2157 = 11^2 cdot mboxBIG
2180 = 7^2 113 cdot mboxBIG
2222 = 7^3 47 cdot mboxBIG
2246 = 7 17 41^2 cdot mboxBIG
2264 = 7^2 cdot mboxBIG
2267 = 11^2 cdot mboxBIG
2306 = 7^2 887 cdot mboxBIG
2340 = 127^2 cdot mboxBIG
2348 = 7^2 191 cdot mboxBIG
2377 = 11^2 359 cdot mboxBIG
2390 = 7^2 17^2 23 431 cdot mboxBIG
2432 = 7^2 cdot mboxBIG
2474 = 7^2 cdot mboxBIG
2487 = 11^2 179 cdot mboxBIG
2516 = 7^3 113 cdot mboxBIG
2531 = 1021^2 cdot mboxBIG
2558 = 7^2 cdot mboxBIG
2597 = 11^2 cdot mboxBIG
jagy@phobeusjunior:~$
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
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1
$begingroup$
@OscarLanzi I've been seeing a dermatologist also.
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– Will Jagy
Mar 20 at 1:14
1
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It is hardly supprising, since 23 divides 3^11-2^11, that is, 23 is a sevenite of 3/2.
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– wendy.krieger
Mar 20 at 9:33
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Not just $23|(3^11-2^11)$. Also $23^2|(3^11-2^11)$. The latter is what ultimately gets you.
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– Oscar Lanzi
Mar 20 at 12:41
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I was half surprised. I was exploring Sophie Germain (23, 47,...) and Mersenne Primes (suposedly squarefree) when I noticed the factorisation (That I thought squarefree too).
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– Collag3n
Mar 20 at 13:44
$begingroup$
I wonder what would be the special number "23" for $5^n-2^n-1$, $7^n-2^n-1$, ...
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– Collag3n
Mar 20 at 18:17
|
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$begingroup$
We can show that there are multiples of $49$ using elementary modular arithmetic techniques.
Let $p$ be a prime greater than $3$ (why?), and seek values of $n$ for which
$$3^nequiv 2^n-1bmod p^2$$
Multiply by $(3^-1)^n-1$ getting
$$3equiv (2×3^-1)^n-1bmod p^2$$
Let us first try $p=5$. We die because the left side is a nonquadratic residue $bmod p=5$ and the right side, with $2×3^-1equiv 4bmod 5$, is a quadratic residue.
Fortunately, for $p=7$ we avoid this contradiction because $2×3^-1$ is nonquadratic $bmod 7$ thus also nonquadratic $bmod 49$. We then have
$$3equiv 17^n-1bmod 49$$
where $17equiv 3bmod 7$ is a primitive root in the group of units $bmod 49$ (the only nonprimitive root $bmod 49$ congruent to $3bmod 7$ is $31$), therefore this equation must have positive whole number solutions for $n$.
Will Jagy has identified the minimal solution as $n=38$, so let us check this case $bmod 49$. Since units give $1$ when raised to the power of $42$, we may render
$$3^38equiv (3^-1)^4equiv 33^4equiv 11^2equiv 121equivcolorblue23bmod 49$$
And
$$2^37equiv (2^-1)^5equiv 25^5equiv 25×(-12)^2equiv 3600equiv colorblue23bmod 49$$
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
$endgroup$
$begingroup$
Oscar, I realized I had a quick way to find squares of small primes, as I write my own factoring-related routines to get quick factors if needed: 712 +++ = 271^2 cdot mboxBIG 1588 +++ = 73 191^2 cdot mboxBIG 2340 +++ = 127^2 cdot mboxBIG 2531 +++ = 1021^2 cdot mboxBIG
$endgroup$
– Will Jagy
Mar 20 at 15:46
add a comment |
$begingroup$
We can show that there are multiples of $49$ using elementary modular arithmetic techniques.
Let $p$ be a prime greater than $3$ (why?), and seek values of $n$ for which
$$3^nequiv 2^n-1bmod p^2$$
Multiply by $(3^-1)^n-1$ getting
$$3equiv (2×3^-1)^n-1bmod p^2$$
Let us first try $p=5$. We die because the left side is a nonquadratic residue $bmod p=5$ and the right side, with $2×3^-1equiv 4bmod 5$, is a quadratic residue.
Fortunately, for $p=7$ we avoid this contradiction because $2×3^-1$ is nonquadratic $bmod 7$ thus also nonquadratic $bmod 49$. We then have
$$3equiv 17^n-1bmod 49$$
where $17equiv 3bmod 7$ is a primitive root in the group of units $bmod 49$ (the only nonprimitive root $bmod 49$ congruent to $3bmod 7$ is $31$), therefore this equation must have positive whole number solutions for $n$.
Will Jagy has identified the minimal solution as $n=38$, so let us check this case $bmod 49$. Since units give $1$ when raised to the power of $42$, we may render
$$3^38equiv (3^-1)^4equiv 33^4equiv 11^2equiv 121equivcolorblue23bmod 49$$
And
$$2^37equiv (2^-1)^5equiv 25^5equiv 25×(-12)^2equiv 3600equiv colorblue23bmod 49$$
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
$endgroup$
$begingroup$
Oscar, I realized I had a quick way to find squares of small primes, as I write my own factoring-related routines to get quick factors if needed: 712 +++ = 271^2 cdot mboxBIG 1588 +++ = 73 191^2 cdot mboxBIG 2340 +++ = 127^2 cdot mboxBIG 2531 +++ = 1021^2 cdot mboxBIG
$endgroup$
– Will Jagy
Mar 20 at 15:46
add a comment |
$begingroup$
We can show that there are multiples of $49$ using elementary modular arithmetic techniques.
Let $p$ be a prime greater than $3$ (why?), and seek values of $n$ for which
$$3^nequiv 2^n-1bmod p^2$$
Multiply by $(3^-1)^n-1$ getting
$$3equiv (2×3^-1)^n-1bmod p^2$$
Let us first try $p=5$. We die because the left side is a nonquadratic residue $bmod p=5$ and the right side, with $2×3^-1equiv 4bmod 5$, is a quadratic residue.
Fortunately, for $p=7$ we avoid this contradiction because $2×3^-1$ is nonquadratic $bmod 7$ thus also nonquadratic $bmod 49$. We then have
$$3equiv 17^n-1bmod 49$$
where $17equiv 3bmod 7$ is a primitive root in the group of units $bmod 49$ (the only nonprimitive root $bmod 49$ congruent to $3bmod 7$ is $31$), therefore this equation must have positive whole number solutions for $n$.
Will Jagy has identified the minimal solution as $n=38$, so let us check this case $bmod 49$. Since units give $1$ when raised to the power of $42$, we may render
$$3^38equiv (3^-1)^4equiv 33^4equiv 11^2equiv 121equivcolorblue23bmod 49$$
And
$$2^37equiv (2^-1)^5equiv 25^5equiv 25×(-12)^2equiv 3600equiv colorblue23bmod 49$$
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
$endgroup$
We can show that there are multiples of $49$ using elementary modular arithmetic techniques.
Let $p$ be a prime greater than $3$ (why?), and seek values of $n$ for which
$$3^nequiv 2^n-1bmod p^2$$
Multiply by $(3^-1)^n-1$ getting
$$3equiv (2×3^-1)^n-1bmod p^2$$
Let us first try $p=5$. We die because the left side is a nonquadratic residue $bmod p=5$ and the right side, with $2×3^-1equiv 4bmod 5$, is a quadratic residue.
Fortunately, for $p=7$ we avoid this contradiction because $2×3^-1$ is nonquadratic $bmod 7$ thus also nonquadratic $bmod 49$. We then have
$$3equiv 17^n-1bmod 49$$
where $17equiv 3bmod 7$ is a primitive root in the group of units $bmod 49$ (the only nonprimitive root $bmod 49$ congruent to $3bmod 7$ is $31$), therefore this equation must have positive whole number solutions for $n$.
Will Jagy has identified the minimal solution as $n=38$, so let us check this case $bmod 49$. Since units give $1$ when raised to the power of $42$, we may render
$$3^38equiv (3^-1)^4equiv 33^4equiv 11^2equiv 121equivcolorblue23bmod 49$$
And
$$2^37equiv (2^-1)^5equiv 25^5equiv 25×(-12)^2equiv 3600equiv colorblue23bmod 49$$
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
edited Mar 20 at 15:08
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
answered Mar 20 at 0:46
Oscar LanziOscar Lanzi
13.4k12136
13.4k12136
$begingroup$
Oscar, I realized I had a quick way to find squares of small primes, as I write my own factoring-related routines to get quick factors if needed: 712 +++ = 271^2 cdot mboxBIG 1588 +++ = 73 191^2 cdot mboxBIG 2340 +++ = 127^2 cdot mboxBIG 2531 +++ = 1021^2 cdot mboxBIG
$endgroup$
– Will Jagy
Mar 20 at 15:46
add a comment |
$begingroup$
Oscar, I realized I had a quick way to find squares of small primes, as I write my own factoring-related routines to get quick factors if needed: 712 +++ = 271^2 cdot mboxBIG 1588 +++ = 73 191^2 cdot mboxBIG 2340 +++ = 127^2 cdot mboxBIG 2531 +++ = 1021^2 cdot mboxBIG
$endgroup$
– Will Jagy
Mar 20 at 15:46
$begingroup$
Oscar, I realized I had a quick way to find squares of small primes, as I write my own factoring-related routines to get quick factors if needed: 712 +++ = 271^2 cdot mboxBIG 1588 +++ = 73 191^2 cdot mboxBIG 2340 +++ = 127^2 cdot mboxBIG 2531 +++ = 1021^2 cdot mboxBIG
$endgroup$
– Will Jagy
Mar 20 at 15:46
$begingroup$
Oscar, I realized I had a quick way to find squares of small primes, as I write my own factoring-related routines to get quick factors if needed: 712 +++ = 271^2 cdot mboxBIG 1588 +++ = 73 191^2 cdot mboxBIG 2340 +++ = 127^2 cdot mboxBIG 2531 +++ = 1021^2 cdot mboxBIG
$endgroup$
– Will Jagy
Mar 20 at 15:46
add a comment |
$begingroup$
Well, I posted too quickly.....for $n=67$ it seems we have 2 factor $11$
answered Mar 19 at 20:49
Collag3nCollag3n
784211
$endgroup$
4
$begingroup$
Also, $7^2 mid 3^38 - 2^37$.
$endgroup$
– Misha Lavrov
Mar 19 at 20:50
1
$begingroup$
Yes, That's because I missed that one that i posted....too quickly
$endgroup$
– Collag3n
Mar 20 at 8:38
add a comment |
$begingroup$
Well, I posted too quickly.....for $n=67$ it seems we have 2 factor $11$
answered Mar 19 at 20:49
Collag3nCollag3n
784211
$endgroup$
4
$begingroup$
Also, $7^2 mid 3^38 - 2^37$.
$endgroup$
– Misha Lavrov
Mar 19 at 20:50
1
$begingroup$
Yes, That's because I missed that one that i posted....too quickly
$endgroup$
– Collag3n
Mar 20 at 8:38
add a comment |
$begingroup$
Well, I posted too quickly.....for $n=67$ it seems we have 2 factor $11$
answered Mar 19 at 20:49
Collag3nCollag3n
784211
$endgroup$
Well, I posted too quickly.....for $n=67$ it seems we have 2 factor $11$
answered Mar 19 at 20:49
Collag3nCollag3n
784211
answered Mar 19 at 20:49
Collag3nCollag3n
784211
answered Mar 19 at 20:49
Collag3nCollag3n
784211
answered Mar 19 at 20:49
Collag3nCollag3n
784211
784211
4
$begingroup$
Also, $7^2 mid 3^38 - 2^37$.
$endgroup$
– Misha Lavrov
Mar 19 at 20:50
1
$begingroup$
Yes, That's because I missed that one that i posted....too quickly
$endgroup$
– Collag3n
Mar 20 at 8:38
add a comment |
4
$begingroup$
Also, $7^2 mid 3^38 - 2^37$.
$endgroup$
– Misha Lavrov
Mar 19 at 20:50
1
$begingroup$
Yes, That's because I missed that one that i posted....too quickly
$endgroup$
– Collag3n
Mar 20 at 8:38
4
4
$begingroup$
Also, $7^2 mid 3^38 - 2^37$.
$endgroup$
– Misha Lavrov
Mar 19 at 20:50
$begingroup$
Also, $7^2 mid 3^38 - 2^37$.
$endgroup$
– Misha Lavrov
Mar 19 at 20:50
1
1
$begingroup$
Yes, That's because I missed that one that i posted....too quickly
$endgroup$
– Collag3n
Mar 20 at 8:38
$begingroup$
Yes, That's because I missed that one that i posted....too quickly
$endgroup$
– Collag3n
Mar 20 at 8:38
add a comment |
$begingroup$
At the very end, I put a long list of these. The great majority of squared primes are below $100,$ but I did catch a few larger, so far $127, 191, 271, 1021 : ;$
712 +++ = 271^2 cdot mboxBIG
1588 +++ = 73 191^2 cdot mboxBIG
2340 +++ = 127^2 cdot mboxBIG
2531 +++ = 1021^2 cdot mboxBIG
There is a way to deal with this without doing genuine factoring. For example, your number is divisible by $49 = 7^2$ whenever $n = 42 k + 38.$
2 7 prime 7
8 6433 prime 7
14 4774777 prime 7
20 3486260113 prime 7
38 1350851580234038617 prime squared 49
80 147808829414345318853173402891795944513 prime squared 49
122 16173092699229880893715960009594875525837473033720099268457 prime squared 49
Your number is divisible by $121 = 11^2$ whenever $n = 110 k + 67.$
7 2123 prime 11
17 129074627 prime 11
27 7625530376123 prime 11
67 92709463147824050109467087204123 prime squared 121
177 2821383260958014531084804730393073172748132970923952481977527762896658545213494562627 prime squared 121
287 85861822891966292897565943276292392423908891501494514127947492094325821620603901184289283467528859878643948705742648123768724313989998123 prime squared 121
Your number is divisible by $289 = 17^2$ whenever $n = 272 k + 214.$
6 697 prime 17
22 31378962457 prime 17
38 1350851580234038617 prime 17
214 1270423474759653869629541561076150845942627861345583970679777076713806668073848513969400550694997546777 prime squared 289
486 7602033756829688179535612101927342434798006222913345882096671718462026450847558385638299237091029898106108915679642725019874527596206183615695170393375601813754958083630064304945006176793537681250116409274581709738622832815408017497 prime squared 289
758 45489491014727452017657094699993344217699387580459634625293727208970468768085650812024804178818092172750636489054819257623093543090018879398677204096366573883113560178809969964755425161715086488029060167818542608680433377185439106005853203184618487801892144052367301608312676367936439310746339829776474186019806821915640418802179770139744725661919759420980369817 prime squared 289
Surprising, your number is often divisible by $23$ but never by $529 = 23^2.$ Go Figure.
3 23 prime 23
14 4774777 prime 23
25 847271832227 prime 23
36 150094600937260753 prime 23
47 26588814288588759110123 prime 23
58 4710128697102129646845747817 prime 23
NO 23 SQUARED
Your number is divisible by $961 = 31^2$ whenever $n = 930 k + 828.$
18 387289417 prime 31
48 79766442936135021508033 prime 31
78 16423203268260507030504015972062417017 prime 31
828 prime squared 961
1758 prime squared 961
2688 prime squared 961
3618 prime squared 961
4548 prime squared 961
Your number is divisible by $1369 = 37^2$ whenever $n = 1332 k + 383.$
23 94138984523 prime 37
59 14130386091450504128613099323 prime 37
95 2120895147045314099684568958946760345244084523 prime 37
383 prime squared 1369
1715 prime squared 1369
3047 prime squared 1369
4379 prime squared 1369
5711 prime squared 1369
I also did a bounded factoring: given one of these numbers, use trial division with primes $p < 1200.$ I did catch a $1021^2$ this way..
jagy@phobeusjunior:~$ ./mse | grep "^"
38 = 7^2 17 cdot mboxBIG
67 = 11^2 cdot mboxBIG
80 = 7^2 23 607 cdot mboxBIG
122 = 7^2 137 599 cdot mboxBIG
164 = 7^3 113 cdot mboxBIG
177 = 11^2 cdot mboxBIG
206 = 7^2 41 cdot mboxBIG
214 = 17^2 cdot mboxBIG
248 = 7^2 cdot mboxBIG
287 = 11^2 cdot mboxBIG
290 = 7^2 47 809 1033 cdot mboxBIG
332 = 7^2 1193 cdot mboxBIG
374 = 7^2 17 1087 cdot mboxBIG
383 = 37^2 cdot mboxBIG
397 = 11^3 cdot mboxBIG
416 = 7^2 233 cdot mboxBIG
458 = 7^3 439 cdot mboxBIG
486 = 17^2 41 cdot mboxBIG
500 = 7^2 113 cdot mboxBIG
507 = 11^2 83 cdot mboxBIG
508 = 73^2 cdot mboxBIG
542 = 7^2 23 cdot mboxBIG
584 = 7^2 431 cdot mboxBIG
606 = 41^2 cdot mboxBIG
617 = 11^2 cdot mboxBIG
626 = 7^2 cdot mboxBIG
668 = 7^2 cdot mboxBIG
710 = 7^2 17 911 cdot mboxBIG
712 = 271^2 cdot mboxBIG
727 = 11^2 47 cdot mboxBIG
752 = 7^3 89 cdot mboxBIG
758 = 7 17^3 cdot mboxBIG
794 = 7^2 cdot mboxBIG
828 = 23 31^2 127 191 cdot mboxBIG
836 = 7^2 113 cdot mboxBIG
837 = 11^2 683 cdot mboxBIG
878 = 7^2 cdot mboxBIG
920 = 7^2 cdot mboxBIG
947 = 11^2 983 cdot mboxBIG
957 = 11 47^2 229 cdot mboxBIG
962 = 7^2 cdot mboxBIG
1004 = 7^2 23 937 cdot mboxBIG
1030 = 17^2 151 cdot mboxBIG
1046 = 7^3 17 41 cdot mboxBIG
1057 = 11^2 59 431 cdot mboxBIG
1088 = 7^2 cdot mboxBIG
1130 = 7^2 cdot mboxBIG
1167 = 11^2 cdot mboxBIG
1172 = 7^2 113 cdot mboxBIG
1214 = 7^2 569 cdot mboxBIG
1256 = 7^2 47 cdot mboxBIG
1277 = 11^2 cdot mboxBIG
1298 = 7^2 cdot mboxBIG
1302 = 17^2 47 223 263 cdot mboxBIG
1340 = 7^3 cdot mboxBIG
1382 = 7^2 17 cdot mboxBIG
1387 = 11^2 cdot mboxBIG
1424 = 7^2 479 cdot mboxBIG
1466 = 7^2 23 cdot mboxBIG
1491 = 83^2 157 cdot mboxBIG
1497 = 11^2 433 cdot mboxBIG
1508 = 7^2 113 cdot mboxBIG
1550 = 7^2 727 cdot mboxBIG
1574 = 7 17^2 cdot mboxBIG
1588 = 73 191^2 cdot mboxBIG
1592 = 7^2 cdot mboxBIG
1607 = 11^3 37 167 cdot mboxBIG
1634 = 7^5 cdot mboxBIG
1676 = 7^2 cdot mboxBIG
1715 = 37^2 587 cdot mboxBIG
1717 = 11^2 1117 cdot mboxBIG
1718 = 7^2 17 cdot mboxBIG
1758 = 31^2 cdot mboxBIG
1760 = 7^2 cdot mboxBIG
1802 = 7^2 cdot mboxBIG
1827 = 11^2 cdot mboxBIG
1844 = 7^2 113 919 cdot mboxBIG
1846 = 17^2 41 cdot mboxBIG
1886 = 7^2 41 863 cdot mboxBIG
1928 = 7^3 23 cdot mboxBIG
1937 = 11^2 cdot mboxBIG
1970 = 7^2 cdot mboxBIG
2012 = 7^2 cdot mboxBIG
2038 = 17 23 47^2 cdot mboxBIG
2047 = 11^2 cdot mboxBIG
2054 = 7^2 17 cdot mboxBIG
2096 = 7^2 cdot mboxBIG
2118 = 17^2 31 cdot mboxBIG
2138 = 7^2 cdot mboxBIG
2157 = 11^2 cdot mboxBIG
2180 = 7^2 113 cdot mboxBIG
2222 = 7^3 47 cdot mboxBIG
2246 = 7 17 41^2 cdot mboxBIG
2264 = 7^2 cdot mboxBIG
2267 = 11^2 cdot mboxBIG
2306 = 7^2 887 cdot mboxBIG
2340 = 127^2 cdot mboxBIG
2348 = 7^2 191 cdot mboxBIG
2377 = 11^2 359 cdot mboxBIG
2390 = 7^2 17^2 23 431 cdot mboxBIG
2432 = 7^2 cdot mboxBIG
2474 = 7^2 cdot mboxBIG
2487 = 11^2 179 cdot mboxBIG
2516 = 7^3 113 cdot mboxBIG
2531 = 1021^2 cdot mboxBIG
2558 = 7^2 cdot mboxBIG
2597 = 11^2 cdot mboxBIG
jagy@phobeusjunior:~$
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
$endgroup$
1
$begingroup$
@OscarLanzi I've been seeing a dermatologist also.
$endgroup$
– Will Jagy
Mar 20 at 1:14
1
$begingroup$
It is hardly supprising, since 23 divides 3^11-2^11, that is, 23 is a sevenite of 3/2.
$endgroup$
– wendy.krieger
Mar 20 at 9:33
$begingroup$
Not just $23|(3^11-2^11)$. Also $23^2|(3^11-2^11)$. The latter is what ultimately gets you.
$endgroup$
– Oscar Lanzi
Mar 20 at 12:41
$begingroup$
I was half surprised. I was exploring Sophie Germain (23, 47,...) and Mersenne Primes (suposedly squarefree) when I noticed the factorisation (That I thought squarefree too).
$endgroup$
– Collag3n
Mar 20 at 13:44
$begingroup$
I wonder what would be the special number "23" for $5^n-2^n-1$, $7^n-2^n-1$, ...
$endgroup$
– Collag3n
Mar 20 at 18:17
|
show 1 more comment
$begingroup$
At the very end, I put a long list of these. The great majority of squared primes are below $100,$ but I did catch a few larger, so far $127, 191, 271, 1021 : ;$
712 +++ = 271^2 cdot mboxBIG
1588 +++ = 73 191^2 cdot mboxBIG
2340 +++ = 127^2 cdot mboxBIG
2531 +++ = 1021^2 cdot mboxBIG
There is a way to deal with this without doing genuine factoring. For example, your number is divisible by $49 = 7^2$ whenever $n = 42 k + 38.$
2 7 prime 7
8 6433 prime 7
14 4774777 prime 7
20 3486260113 prime 7
38 1350851580234038617 prime squared 49
80 147808829414345318853173402891795944513 prime squared 49
122 16173092699229880893715960009594875525837473033720099268457 prime squared 49
Your number is divisible by $121 = 11^2$ whenever $n = 110 k + 67.$
7 2123 prime 11
17 129074627 prime 11
27 7625530376123 prime 11
67 92709463147824050109467087204123 prime squared 121
177 2821383260958014531084804730393073172748132970923952481977527762896658545213494562627 prime squared 121
287 85861822891966292897565943276292392423908891501494514127947492094325821620603901184289283467528859878643948705742648123768724313989998123 prime squared 121
Your number is divisible by $289 = 17^2$ whenever $n = 272 k + 214.$
6 697 prime 17
22 31378962457 prime 17
38 1350851580234038617 prime 17
214 1270423474759653869629541561076150845942627861345583970679777076713806668073848513969400550694997546777 prime squared 289
486 7602033756829688179535612101927342434798006222913345882096671718462026450847558385638299237091029898106108915679642725019874527596206183615695170393375601813754958083630064304945006176793537681250116409274581709738622832815408017497 prime squared 289
758 45489491014727452017657094699993344217699387580459634625293727208970468768085650812024804178818092172750636489054819257623093543090018879398677204096366573883113560178809969964755425161715086488029060167818542608680433377185439106005853203184618487801892144052367301608312676367936439310746339829776474186019806821915640418802179770139744725661919759420980369817 prime squared 289
Surprising, your number is often divisible by $23$ but never by $529 = 23^2.$ Go Figure.
3 23 prime 23
14 4774777 prime 23
25 847271832227 prime 23
36 150094600937260753 prime 23
47 26588814288588759110123 prime 23
58 4710128697102129646845747817 prime 23
NO 23 SQUARED
Your number is divisible by $961 = 31^2$ whenever $n = 930 k + 828.$
18 387289417 prime 31
48 79766442936135021508033 prime 31
78 16423203268260507030504015972062417017 prime 31
828 prime squared 961
1758 prime squared 961
2688 prime squared 961
3618 prime squared 961
4548 prime squared 961
Your number is divisible by $1369 = 37^2$ whenever $n = 1332 k + 383.$
23 94138984523 prime 37
59 14130386091450504128613099323 prime 37
95 2120895147045314099684568958946760345244084523 prime 37
383 prime squared 1369
1715 prime squared 1369
3047 prime squared 1369
4379 prime squared 1369
5711 prime squared 1369
I also did a bounded factoring: given one of these numbers, use trial division with primes $p < 1200.$ I did catch a $1021^2$ this way..
jagy@phobeusjunior:~$ ./mse | grep "^"
38 = 7^2 17 cdot mboxBIG
67 = 11^2 cdot mboxBIG
80 = 7^2 23 607 cdot mboxBIG
122 = 7^2 137 599 cdot mboxBIG
164 = 7^3 113 cdot mboxBIG
177 = 11^2 cdot mboxBIG
206 = 7^2 41 cdot mboxBIG
214 = 17^2 cdot mboxBIG
248 = 7^2 cdot mboxBIG
287 = 11^2 cdot mboxBIG
290 = 7^2 47 809 1033 cdot mboxBIG
332 = 7^2 1193 cdot mboxBIG
374 = 7^2 17 1087 cdot mboxBIG
383 = 37^2 cdot mboxBIG
397 = 11^3 cdot mboxBIG
416 = 7^2 233 cdot mboxBIG
458 = 7^3 439 cdot mboxBIG
486 = 17^2 41 cdot mboxBIG
500 = 7^2 113 cdot mboxBIG
507 = 11^2 83 cdot mboxBIG
508 = 73^2 cdot mboxBIG
542 = 7^2 23 cdot mboxBIG
584 = 7^2 431 cdot mboxBIG
606 = 41^2 cdot mboxBIG
617 = 11^2 cdot mboxBIG
626 = 7^2 cdot mboxBIG
668 = 7^2 cdot mboxBIG
710 = 7^2 17 911 cdot mboxBIG
712 = 271^2 cdot mboxBIG
727 = 11^2 47 cdot mboxBIG
752 = 7^3 89 cdot mboxBIG
758 = 7 17^3 cdot mboxBIG
794 = 7^2 cdot mboxBIG
828 = 23 31^2 127 191 cdot mboxBIG
836 = 7^2 113 cdot mboxBIG
837 = 11^2 683 cdot mboxBIG
878 = 7^2 cdot mboxBIG
920 = 7^2 cdot mboxBIG
947 = 11^2 983 cdot mboxBIG
957 = 11 47^2 229 cdot mboxBIG
962 = 7^2 cdot mboxBIG
1004 = 7^2 23 937 cdot mboxBIG
1030 = 17^2 151 cdot mboxBIG
1046 = 7^3 17 41 cdot mboxBIG
1057 = 11^2 59 431 cdot mboxBIG
1088 = 7^2 cdot mboxBIG
1130 = 7^2 cdot mboxBIG
1167 = 11^2 cdot mboxBIG
1172 = 7^2 113 cdot mboxBIG
1214 = 7^2 569 cdot mboxBIG
1256 = 7^2 47 cdot mboxBIG
1277 = 11^2 cdot mboxBIG
1298 = 7^2 cdot mboxBIG
1302 = 17^2 47 223 263 cdot mboxBIG
1340 = 7^3 cdot mboxBIG
1382 = 7^2 17 cdot mboxBIG
1387 = 11^2 cdot mboxBIG
1424 = 7^2 479 cdot mboxBIG
1466 = 7^2 23 cdot mboxBIG
1491 = 83^2 157 cdot mboxBIG
1497 = 11^2 433 cdot mboxBIG
1508 = 7^2 113 cdot mboxBIG
1550 = 7^2 727 cdot mboxBIG
1574 = 7 17^2 cdot mboxBIG
1588 = 73 191^2 cdot mboxBIG
1592 = 7^2 cdot mboxBIG
1607 = 11^3 37 167 cdot mboxBIG
1634 = 7^5 cdot mboxBIG
1676 = 7^2 cdot mboxBIG
1715 = 37^2 587 cdot mboxBIG
1717 = 11^2 1117 cdot mboxBIG
1718 = 7^2 17 cdot mboxBIG
1758 = 31^2 cdot mboxBIG
1760 = 7^2 cdot mboxBIG
1802 = 7^2 cdot mboxBIG
1827 = 11^2 cdot mboxBIG
1844 = 7^2 113 919 cdot mboxBIG
1846 = 17^2 41 cdot mboxBIG
1886 = 7^2 41 863 cdot mboxBIG
1928 = 7^3 23 cdot mboxBIG
1937 = 11^2 cdot mboxBIG
1970 = 7^2 cdot mboxBIG
2012 = 7^2 cdot mboxBIG
2038 = 17 23 47^2 cdot mboxBIG
2047 = 11^2 cdot mboxBIG
2054 = 7^2 17 cdot mboxBIG
2096 = 7^2 cdot mboxBIG
2118 = 17^2 31 cdot mboxBIG
2138 = 7^2 cdot mboxBIG
2157 = 11^2 cdot mboxBIG
2180 = 7^2 113 cdot mboxBIG
2222 = 7^3 47 cdot mboxBIG
2246 = 7 17 41^2 cdot mboxBIG
2264 = 7^2 cdot mboxBIG
2267 = 11^2 cdot mboxBIG
2306 = 7^2 887 cdot mboxBIG
2340 = 127^2 cdot mboxBIG
2348 = 7^2 191 cdot mboxBIG
2377 = 11^2 359 cdot mboxBIG
2390 = 7^2 17^2 23 431 cdot mboxBIG
2432 = 7^2 cdot mboxBIG
2474 = 7^2 cdot mboxBIG
2487 = 11^2 179 cdot mboxBIG
2516 = 7^3 113 cdot mboxBIG
2531 = 1021^2 cdot mboxBIG
2558 = 7^2 cdot mboxBIG
2597 = 11^2 cdot mboxBIG
jagy@phobeusjunior:~$
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
$endgroup$
1
$begingroup$
@OscarLanzi I've been seeing a dermatologist also.
$endgroup$
– Will Jagy
Mar 20 at 1:14
1
$begingroup$
It is hardly supprising, since 23 divides 3^11-2^11, that is, 23 is a sevenite of 3/2.
$endgroup$
– wendy.krieger
Mar 20 at 9:33
$begingroup$
Not just $23|(3^11-2^11)$. Also $23^2|(3^11-2^11)$. The latter is what ultimately gets you.
$endgroup$
– Oscar Lanzi
Mar 20 at 12:41
$begingroup$
I was half surprised. I was exploring Sophie Germain (23, 47,...) and Mersenne Primes (suposedly squarefree) when I noticed the factorisation (That I thought squarefree too).
$endgroup$
– Collag3n
Mar 20 at 13:44
$begingroup$
I wonder what would be the special number "23" for $5^n-2^n-1$, $7^n-2^n-1$, ...
$endgroup$
– Collag3n
Mar 20 at 18:17
|
show 1 more comment
$begingroup$
At the very end, I put a long list of these. The great majority of squared primes are below $100,$ but I did catch a few larger, so far $127, 191, 271, 1021 : ;$
712 +++ = 271^2 cdot mboxBIG
1588 +++ = 73 191^2 cdot mboxBIG
2340 +++ = 127^2 cdot mboxBIG
2531 +++ = 1021^2 cdot mboxBIG
There is a way to deal with this without doing genuine factoring. For example, your number is divisible by $49 = 7^2$ whenever $n = 42 k + 38.$
2 7 prime 7
8 6433 prime 7
14 4774777 prime 7
20 3486260113 prime 7
38 1350851580234038617 prime squared 49
80 147808829414345318853173402891795944513 prime squared 49
122 16173092699229880893715960009594875525837473033720099268457 prime squared 49
Your number is divisible by $121 = 11^2$ whenever $n = 110 k + 67.$
7 2123 prime 11
17 129074627 prime 11
27 7625530376123 prime 11
67 92709463147824050109467087204123 prime squared 121
177 2821383260958014531084804730393073172748132970923952481977527762896658545213494562627 prime squared 121
287 85861822891966292897565943276292392423908891501494514127947492094325821620603901184289283467528859878643948705742648123768724313989998123 prime squared 121
Your number is divisible by $289 = 17^2$ whenever $n = 272 k + 214.$
6 697 prime 17
22 31378962457 prime 17
38 1350851580234038617 prime 17
214 1270423474759653869629541561076150845942627861345583970679777076713806668073848513969400550694997546777 prime squared 289
486 7602033756829688179535612101927342434798006222913345882096671718462026450847558385638299237091029898106108915679642725019874527596206183615695170393375601813754958083630064304945006176793537681250116409274581709738622832815408017497 prime squared 289
758 45489491014727452017657094699993344217699387580459634625293727208970468768085650812024804178818092172750636489054819257623093543090018879398677204096366573883113560178809969964755425161715086488029060167818542608680433377185439106005853203184618487801892144052367301608312676367936439310746339829776474186019806821915640418802179770139744725661919759420980369817 prime squared 289
Surprising, your number is often divisible by $23$ but never by $529 = 23^2.$ Go Figure.
3 23 prime 23
14 4774777 prime 23
25 847271832227 prime 23
36 150094600937260753 prime 23
47 26588814288588759110123 prime 23
58 4710128697102129646845747817 prime 23
NO 23 SQUARED
Your number is divisible by $961 = 31^2$ whenever $n = 930 k + 828.$
18 387289417 prime 31
48 79766442936135021508033 prime 31
78 16423203268260507030504015972062417017 prime 31
828 prime squared 961
1758 prime squared 961
2688 prime squared 961
3618 prime squared 961
4548 prime squared 961
Your number is divisible by $1369 = 37^2$ whenever $n = 1332 k + 383.$
23 94138984523 prime 37
59 14130386091450504128613099323 prime 37
95 2120895147045314099684568958946760345244084523 prime 37
383 prime squared 1369
1715 prime squared 1369
3047 prime squared 1369
4379 prime squared 1369
5711 prime squared 1369
I also did a bounded factoring: given one of these numbers, use trial division with primes $p < 1200.$ I did catch a $1021^2$ this way..
jagy@phobeusjunior:~$ ./mse | grep "^"
38 = 7^2 17 cdot mboxBIG
67 = 11^2 cdot mboxBIG
80 = 7^2 23 607 cdot mboxBIG
122 = 7^2 137 599 cdot mboxBIG
164 = 7^3 113 cdot mboxBIG
177 = 11^2 cdot mboxBIG
206 = 7^2 41 cdot mboxBIG
214 = 17^2 cdot mboxBIG
248 = 7^2 cdot mboxBIG
287 = 11^2 cdot mboxBIG
290 = 7^2 47 809 1033 cdot mboxBIG
332 = 7^2 1193 cdot mboxBIG
374 = 7^2 17 1087 cdot mboxBIG
383 = 37^2 cdot mboxBIG
397 = 11^3 cdot mboxBIG
416 = 7^2 233 cdot mboxBIG
458 = 7^3 439 cdot mboxBIG
486 = 17^2 41 cdot mboxBIG
500 = 7^2 113 cdot mboxBIG
507 = 11^2 83 cdot mboxBIG
508 = 73^2 cdot mboxBIG
542 = 7^2 23 cdot mboxBIG
584 = 7^2 431 cdot mboxBIG
606 = 41^2 cdot mboxBIG
617 = 11^2 cdot mboxBIG
626 = 7^2 cdot mboxBIG
668 = 7^2 cdot mboxBIG
710 = 7^2 17 911 cdot mboxBIG
712 = 271^2 cdot mboxBIG
727 = 11^2 47 cdot mboxBIG
752 = 7^3 89 cdot mboxBIG
758 = 7 17^3 cdot mboxBIG
794 = 7^2 cdot mboxBIG
828 = 23 31^2 127 191 cdot mboxBIG
836 = 7^2 113 cdot mboxBIG
837 = 11^2 683 cdot mboxBIG
878 = 7^2 cdot mboxBIG
920 = 7^2 cdot mboxBIG
947 = 11^2 983 cdot mboxBIG
957 = 11 47^2 229 cdot mboxBIG
962 = 7^2 cdot mboxBIG
1004 = 7^2 23 937 cdot mboxBIG
1030 = 17^2 151 cdot mboxBIG
1046 = 7^3 17 41 cdot mboxBIG
1057 = 11^2 59 431 cdot mboxBIG
1088 = 7^2 cdot mboxBIG
1130 = 7^2 cdot mboxBIG
1167 = 11^2 cdot mboxBIG
1172 = 7^2 113 cdot mboxBIG
1214 = 7^2 569 cdot mboxBIG
1256 = 7^2 47 cdot mboxBIG
1277 = 11^2 cdot mboxBIG
1298 = 7^2 cdot mboxBIG
1302 = 17^2 47 223 263 cdot mboxBIG
1340 = 7^3 cdot mboxBIG
1382 = 7^2 17 cdot mboxBIG
1387 = 11^2 cdot mboxBIG
1424 = 7^2 479 cdot mboxBIG
1466 = 7^2 23 cdot mboxBIG
1491 = 83^2 157 cdot mboxBIG
1497 = 11^2 433 cdot mboxBIG
1508 = 7^2 113 cdot mboxBIG
1550 = 7^2 727 cdot mboxBIG
1574 = 7 17^2 cdot mboxBIG
1588 = 73 191^2 cdot mboxBIG
1592 = 7^2 cdot mboxBIG
1607 = 11^3 37 167 cdot mboxBIG
1634 = 7^5 cdot mboxBIG
1676 = 7^2 cdot mboxBIG
1715 = 37^2 587 cdot mboxBIG
1717 = 11^2 1117 cdot mboxBIG
1718 = 7^2 17 cdot mboxBIG
1758 = 31^2 cdot mboxBIG
1760 = 7^2 cdot mboxBIG
1802 = 7^2 cdot mboxBIG
1827 = 11^2 cdot mboxBIG
1844 = 7^2 113 919 cdot mboxBIG
1846 = 17^2 41 cdot mboxBIG
1886 = 7^2 41 863 cdot mboxBIG
1928 = 7^3 23 cdot mboxBIG
1937 = 11^2 cdot mboxBIG
1970 = 7^2 cdot mboxBIG
2012 = 7^2 cdot mboxBIG
2038 = 17 23 47^2 cdot mboxBIG
2047 = 11^2 cdot mboxBIG
2054 = 7^2 17 cdot mboxBIG
2096 = 7^2 cdot mboxBIG
2118 = 17^2 31 cdot mboxBIG
2138 = 7^2 cdot mboxBIG
2157 = 11^2 cdot mboxBIG
2180 = 7^2 113 cdot mboxBIG
2222 = 7^3 47 cdot mboxBIG
2246 = 7 17 41^2 cdot mboxBIG
2264 = 7^2 cdot mboxBIG
2267 = 11^2 cdot mboxBIG
2306 = 7^2 887 cdot mboxBIG
2340 = 127^2 cdot mboxBIG
2348 = 7^2 191 cdot mboxBIG
2377 = 11^2 359 cdot mboxBIG
2390 = 7^2 17^2 23 431 cdot mboxBIG
2432 = 7^2 cdot mboxBIG
2474 = 7^2 cdot mboxBIG
2487 = 11^2 179 cdot mboxBIG
2516 = 7^3 113 cdot mboxBIG
2531 = 1021^2 cdot mboxBIG
2558 = 7^2 cdot mboxBIG
2597 = 11^2 cdot mboxBIG
jagy@phobeusjunior:~$
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
$endgroup$
At the very end, I put a long list of these. The great majority of squared primes are below $100,$ but I did catch a few larger, so far $127, 191, 271, 1021 : ;$
712 +++ = 271^2 cdot mboxBIG
1588 +++ = 73 191^2 cdot mboxBIG
2340 +++ = 127^2 cdot mboxBIG
2531 +++ = 1021^2 cdot mboxBIG
There is a way to deal with this without doing genuine factoring. For example, your number is divisible by $49 = 7^2$ whenever $n = 42 k + 38.$
2 7 prime 7
8 6433 prime 7
14 4774777 prime 7
20 3486260113 prime 7
38 1350851580234038617 prime squared 49
80 147808829414345318853173402891795944513 prime squared 49
122 16173092699229880893715960009594875525837473033720099268457 prime squared 49
Your number is divisible by $121 = 11^2$ whenever $n = 110 k + 67.$
7 2123 prime 11
17 129074627 prime 11
27 7625530376123 prime 11
67 92709463147824050109467087204123 prime squared 121
177 2821383260958014531084804730393073172748132970923952481977527762896658545213494562627 prime squared 121
287 85861822891966292897565943276292392423908891501494514127947492094325821620603901184289283467528859878643948705742648123768724313989998123 prime squared 121
Your number is divisible by $289 = 17^2$ whenever $n = 272 k + 214.$
6 697 prime 17
22 31378962457 prime 17
38 1350851580234038617 prime 17
214 1270423474759653869629541561076150845942627861345583970679777076713806668073848513969400550694997546777 prime squared 289
486 7602033756829688179535612101927342434798006222913345882096671718462026450847558385638299237091029898106108915679642725019874527596206183615695170393375601813754958083630064304945006176793537681250116409274581709738622832815408017497 prime squared 289
758 45489491014727452017657094699993344217699387580459634625293727208970468768085650812024804178818092172750636489054819257623093543090018879398677204096366573883113560178809969964755425161715086488029060167818542608680433377185439106005853203184618487801892144052367301608312676367936439310746339829776474186019806821915640418802179770139744725661919759420980369817 prime squared 289
Surprising, your number is often divisible by $23$ but never by $529 = 23^2.$ Go Figure.
3 23 prime 23
14 4774777 prime 23
25 847271832227 prime 23
36 150094600937260753 prime 23
47 26588814288588759110123 prime 23
58 4710128697102129646845747817 prime 23
NO 23 SQUARED
Your number is divisible by $961 = 31^2$ whenever $n = 930 k + 828.$
18 387289417 prime 31
48 79766442936135021508033 prime 31
78 16423203268260507030504015972062417017 prime 31
828 prime squared 961
1758 prime squared 961
2688 prime squared 961
3618 prime squared 961
4548 prime squared 961
Your number is divisible by $1369 = 37^2$ whenever $n = 1332 k + 383.$
23 94138984523 prime 37
59 14130386091450504128613099323 prime 37
95 2120895147045314099684568958946760345244084523 prime 37
383 prime squared 1369
1715 prime squared 1369
3047 prime squared 1369
4379 prime squared 1369
5711 prime squared 1369
I also did a bounded factoring: given one of these numbers, use trial division with primes $p < 1200.$ I did catch a $1021^2$ this way..
jagy@phobeusjunior:~$ ./mse | grep "^"
38 = 7^2 17 cdot mboxBIG
67 = 11^2 cdot mboxBIG
80 = 7^2 23 607 cdot mboxBIG
122 = 7^2 137 599 cdot mboxBIG
164 = 7^3 113 cdot mboxBIG
177 = 11^2 cdot mboxBIG
206 = 7^2 41 cdot mboxBIG
214 = 17^2 cdot mboxBIG
248 = 7^2 cdot mboxBIG
287 = 11^2 cdot mboxBIG
290 = 7^2 47 809 1033 cdot mboxBIG
332 = 7^2 1193 cdot mboxBIG
374 = 7^2 17 1087 cdot mboxBIG
383 = 37^2 cdot mboxBIG
397 = 11^3 cdot mboxBIG
416 = 7^2 233 cdot mboxBIG
458 = 7^3 439 cdot mboxBIG
486 = 17^2 41 cdot mboxBIG
500 = 7^2 113 cdot mboxBIG
507 = 11^2 83 cdot mboxBIG
508 = 73^2 cdot mboxBIG
542 = 7^2 23 cdot mboxBIG
584 = 7^2 431 cdot mboxBIG
606 = 41^2 cdot mboxBIG
617 = 11^2 cdot mboxBIG
626 = 7^2 cdot mboxBIG
668 = 7^2 cdot mboxBIG
710 = 7^2 17 911 cdot mboxBIG
712 = 271^2 cdot mboxBIG
727 = 11^2 47 cdot mboxBIG
752 = 7^3 89 cdot mboxBIG
758 = 7 17^3 cdot mboxBIG
794 = 7^2 cdot mboxBIG
828 = 23 31^2 127 191 cdot mboxBIG
836 = 7^2 113 cdot mboxBIG
837 = 11^2 683 cdot mboxBIG
878 = 7^2 cdot mboxBIG
920 = 7^2 cdot mboxBIG
947 = 11^2 983 cdot mboxBIG
957 = 11 47^2 229 cdot mboxBIG
962 = 7^2 cdot mboxBIG
1004 = 7^2 23 937 cdot mboxBIG
1030 = 17^2 151 cdot mboxBIG
1046 = 7^3 17 41 cdot mboxBIG
1057 = 11^2 59 431 cdot mboxBIG
1088 = 7^2 cdot mboxBIG
1130 = 7^2 cdot mboxBIG
1167 = 11^2 cdot mboxBIG
1172 = 7^2 113 cdot mboxBIG
1214 = 7^2 569 cdot mboxBIG
1256 = 7^2 47 cdot mboxBIG
1277 = 11^2 cdot mboxBIG
1298 = 7^2 cdot mboxBIG
1302 = 17^2 47 223 263 cdot mboxBIG
1340 = 7^3 cdot mboxBIG
1382 = 7^2 17 cdot mboxBIG
1387 = 11^2 cdot mboxBIG
1424 = 7^2 479 cdot mboxBIG
1466 = 7^2 23 cdot mboxBIG
1491 = 83^2 157 cdot mboxBIG
1497 = 11^2 433 cdot mboxBIG
1508 = 7^2 113 cdot mboxBIG
1550 = 7^2 727 cdot mboxBIG
1574 = 7 17^2 cdot mboxBIG
1588 = 73 191^2 cdot mboxBIG
1592 = 7^2 cdot mboxBIG
1607 = 11^3 37 167 cdot mboxBIG
1634 = 7^5 cdot mboxBIG
1676 = 7^2 cdot mboxBIG
1715 = 37^2 587 cdot mboxBIG
1717 = 11^2 1117 cdot mboxBIG
1718 = 7^2 17 cdot mboxBIG
1758 = 31^2 cdot mboxBIG
1760 = 7^2 cdot mboxBIG
1802 = 7^2 cdot mboxBIG
1827 = 11^2 cdot mboxBIG
1844 = 7^2 113 919 cdot mboxBIG
1846 = 17^2 41 cdot mboxBIG
1886 = 7^2 41 863 cdot mboxBIG
1928 = 7^3 23 cdot mboxBIG
1937 = 11^2 cdot mboxBIG
1970 = 7^2 cdot mboxBIG
2012 = 7^2 cdot mboxBIG
2038 = 17 23 47^2 cdot mboxBIG
2047 = 11^2 cdot mboxBIG
2054 = 7^2 17 cdot mboxBIG
2096 = 7^2 cdot mboxBIG
2118 = 17^2 31 cdot mboxBIG
2138 = 7^2 cdot mboxBIG
2157 = 11^2 cdot mboxBIG
2180 = 7^2 113 cdot mboxBIG
2222 = 7^3 47 cdot mboxBIG
2246 = 7 17 41^2 cdot mboxBIG
2264 = 7^2 cdot mboxBIG
2267 = 11^2 cdot mboxBIG
2306 = 7^2 887 cdot mboxBIG
2340 = 127^2 cdot mboxBIG
2348 = 7^2 191 cdot mboxBIG
2377 = 11^2 359 cdot mboxBIG
2390 = 7^2 17^2 23 431 cdot mboxBIG
2432 = 7^2 cdot mboxBIG
2474 = 7^2 cdot mboxBIG
2487 = 11^2 179 cdot mboxBIG
2516 = 7^3 113 cdot mboxBIG
2531 = 1021^2 cdot mboxBIG
2558 = 7^2 cdot mboxBIG
2597 = 11^2 cdot mboxBIG
jagy@phobeusjunior:~$
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
edited Mar 20 at 15:35
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
answered Mar 19 at 23:39
Will JagyWill Jagy
104k5102201
104k5102201
1
$begingroup$
@OscarLanzi I've been seeing a dermatologist also.
$endgroup$
– Will Jagy
Mar 20 at 1:14
1
$begingroup$
It is hardly supprising, since 23 divides 3^11-2^11, that is, 23 is a sevenite of 3/2.
$endgroup$
– wendy.krieger
Mar 20 at 9:33
$begingroup$
Not just $23|(3^11-2^11)$. Also $23^2|(3^11-2^11)$. The latter is what ultimately gets you.
$endgroup$
– Oscar Lanzi
Mar 20 at 12:41
$begingroup$
I was half surprised. I was exploring Sophie Germain (23, 47,...) and Mersenne Primes (suposedly squarefree) when I noticed the factorisation (That I thought squarefree too).
$endgroup$
– Collag3n
Mar 20 at 13:44
$begingroup$
I wonder what would be the special number "23" for $5^n-2^n-1$, $7^n-2^n-1$, ...
$endgroup$
– Collag3n
Mar 20 at 18:17
|
show 1 more comment
1
$begingroup$
@OscarLanzi I've been seeing a dermatologist also.
$endgroup$
– Will Jagy
Mar 20 at 1:14
1
$begingroup$
It is hardly supprising, since 23 divides 3^11-2^11, that is, 23 is a sevenite of 3/2.
$endgroup$
– wendy.krieger
Mar 20 at 9:33
$begingroup$
Not just $23|(3^11-2^11)$. Also $23^2|(3^11-2^11)$. The latter is what ultimately gets you.
$endgroup$
– Oscar Lanzi
Mar 20 at 12:41
$begingroup$
I was half surprised. I was exploring Sophie Germain (23, 47,...) and Mersenne Primes (suposedly squarefree) when I noticed the factorisation (That I thought squarefree too).
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– Collag3n
Mar 20 at 13:44
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I wonder what would be the special number "23" for $5^n-2^n-1$, $7^n-2^n-1$, ...
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– Collag3n
Mar 20 at 18:17
1
1
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@OscarLanzi I've been seeing a dermatologist also.
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– Will Jagy
Mar 20 at 1:14
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@OscarLanzi I've been seeing a dermatologist also.
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– Will Jagy
Mar 20 at 1:14
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1
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It is hardly supprising, since 23 divides 3^11-2^11, that is, 23 is a sevenite of 3/2.
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– wendy.krieger
Mar 20 at 9:33
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It is hardly supprising, since 23 divides 3^11-2^11, that is, 23 is a sevenite of 3/2.
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– wendy.krieger
Mar 20 at 9:33
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Not just $23|(3^11-2^11)$. Also $23^2|(3^11-2^11)$. The latter is what ultimately gets you.
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– Oscar Lanzi
Mar 20 at 12:41
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Not just $23|(3^11-2^11)$. Also $23^2|(3^11-2^11)$. The latter is what ultimately gets you.
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– Oscar Lanzi
Mar 20 at 12:41
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I was half surprised. I was exploring Sophie Germain (23, 47,...) and Mersenne Primes (suposedly squarefree) when I noticed the factorisation (That I thought squarefree too).
$endgroup$
– Collag3n
Mar 20 at 13:44
$begingroup$
I was half surprised. I was exploring Sophie Germain (23, 47,...) and Mersenne Primes (suposedly squarefree) when I noticed the factorisation (That I thought squarefree too).
$endgroup$
– Collag3n
Mar 20 at 13:44
$begingroup$
I wonder what would be the special number "23" for $5^n-2^n-1$, $7^n-2^n-1$, ...
$endgroup$
– Collag3n
Mar 20 at 18:17
$begingroup$
I wonder what would be the special number "23" for $5^n-2^n-1$, $7^n-2^n-1$, ...
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– Collag3n
Mar 20 at 18:17
|
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wolframalpha.com/input/… Nope, but counterexamples seem to be pretty sparse.
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– user113102
Mar 19 at 20:49
1
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Use computer simulations to have an idea about it...
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– Jean Marie
Mar 19 at 20:51
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Thx user113102, I sometimes used wolfram to plot things but didn't thought about using it that way.
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– Collag3n
Mar 19 at 20:54
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Deleted and reposted because I cannot freely edit my OWN comments: For the original problem, $23$ is not unique. There are lots of primes p>3 for which p2 never divides $3^n−2n−1$, the smallest of these actually being $5$. Just to be clear, in the edited question the exponent is now prime, right?
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– Oscar Lanzi
Mar 20 at 22:56
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Yes, sorry, I edited the post. Like Will, I was only looking at prime dividing these numbers, and indeed, I only consider prime exponent in the edited question (since there are clearly squares in the 3 forms for other $n$)
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– Collag3n
Mar 21 at 7:10