Is it true that $(a^2-ab+b^2)(c^2-cd+d^2)=h^2-hk+k^2$ for some coprime $h$ and $k$?$x^2-xy+y^2=n$ how many solutionsFor $a, b$ coprime, if $p geq 5$ is a prime which divides $a^2 - ab + b^2$, then $p equiv 1 pmod6$Showing $aperp b$ and $nne 0$ implies $a+bkperp n$ for some $k$If $x=123456789101112131415161718$, then $xequiv 6pmod16$ and $xequiv 0pmod 6$Are there long arithmetic progressions non-coprime with the given number?Solution to some confusing complex equationProof that the product of primitive Pythagorean hypotenuses is also a primitive Pythagorean hypotenuseIf all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Prove that $a$ and $b$ are coprime whenever $a+b$ and $a-b$ are coprimeMaximal Consecutive Integer SequenceIs the sum of two coprime natural numbers prime?Probability of k random integers being coprimes
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A Trivial Diagnosis
Is it true that $(a^2-ab+b^2)(c^2-cd+d^2)=h^2-hk+k^2$ for some coprime $h$ and $k$?
$x^2-xy+y^2=n$ how many solutionsFor $a, b$ coprime, if $p geq 5$ is a prime which divides $a^2 - ab + b^2$, then $p equiv 1 pmod6$Showing $aperp b$ and $nne 0$ implies $a+bkperp n$ for some $k$If $x=123456789101112131415161718$, then $xequiv 6pmod16$ and $xequiv 0pmod 6$Are there long arithmetic progressions non-coprime with the given number?Solution to some confusing complex equationProof that the product of primitive Pythagorean hypotenuses is also a primitive Pythagorean hypotenuseIf all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Prove that $a$ and $b$ are coprime whenever $a+b$ and $a-b$ are coprimeMaximal Consecutive Integer SequenceIs the sum of two coprime natural numbers prime?Probability of k random integers being coprimes
$begingroup$
Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?
I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!
elementary-number-theory complex-numbers quadratics
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
asked Mar 14 at 15:35
Al TacAl Tac
242
$endgroup$
add a comment |
$begingroup$
Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?
I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!
elementary-number-theory complex-numbers quadratics
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
asked Mar 14 at 15:35
Al TacAl Tac
242
$endgroup$
$begingroup$
The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
$endgroup$
– Dietrich Burde
Mar 14 at 16:20
$begingroup$
Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
$endgroup$
– Al Tac
Mar 14 at 17:33
1
$begingroup$
Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
$endgroup$
– Dietrich Burde
Mar 14 at 17:56
$begingroup$
Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
$endgroup$
– Al Tac
Mar 15 at 8:24
add a comment |
$begingroup$
Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?
I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!
elementary-number-theory complex-numbers quadratics
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
asked Mar 14 at 15:35
Al TacAl Tac
242
$endgroup$
Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true?
I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!
elementary-number-theory complex-numbers quadratics
elementary-number-theory complex-numbers quadratics
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
asked Mar 14 at 15:35
Al TacAl Tac
242
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
asked Mar 14 at 15:35
Al TacAl Tac
242
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
edited Mar 14 at 20:30
Asaf Karagila♦
306k33438769
306k33438769
asked Mar 14 at 15:35
Al TacAl Tac
242
asked Mar 14 at 15:35
Al TacAl Tac
242
asked Mar 14 at 15:35
Al TacAl Tac
242
242
$begingroup$
The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
$endgroup$
– Dietrich Burde
Mar 14 at 16:20
$begingroup$
Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
$endgroup$
– Al Tac
Mar 14 at 17:33
1
$begingroup$
Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
$endgroup$
– Dietrich Burde
Mar 14 at 17:56
$begingroup$
Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
$endgroup$
– Al Tac
Mar 15 at 8:24
add a comment |
$begingroup$
The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
$endgroup$
– Dietrich Burde
Mar 14 at 16:20
$begingroup$
Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
$endgroup$
– Al Tac
Mar 14 at 17:33
1
$begingroup$
Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
$endgroup$
– Dietrich Burde
Mar 14 at 17:56
$begingroup$
Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
$endgroup$
– Al Tac
Mar 15 at 8:24
$begingroup$
The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
$endgroup$
– Dietrich Burde
Mar 14 at 16:20
$begingroup$
The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
$endgroup$
– Dietrich Burde
Mar 14 at 16:20
$begingroup$
Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
$endgroup$
– Al Tac
Mar 14 at 17:33
$begingroup$
Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
$endgroup$
– Al Tac
Mar 14 at 17:33
1
1
$begingroup$
Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
$endgroup$
– Dietrich Burde
Mar 14 at 17:56
$begingroup$
Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
$endgroup$
– Dietrich Burde
Mar 14 at 17:56
$begingroup$
Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
$endgroup$
– Al Tac
Mar 15 at 8:24
$begingroup$
Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
$endgroup$
– Al Tac
Mar 15 at 8:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$
so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$
$$ =(m-nu)(m-nv) = m^2-mn+n^2$$
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
1
$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52
$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
Mar 14 at 19:25
$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
Mar 14 at 19:38
add a comment |
$begingroup$
There is this Identity:
$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$
Hence for:
$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$
$h=(ac+bd)$
$k=(bc-ad)$
$hk=(ac+bd)(bc-ad)$
Condition (c,d)=(2b,b-2a)
For $(a,b,c,d)=(3,7,14,1)$ we get:
$(49^2-49*95+95^2)=(37)*(183)=6771$
answered Mar 14 at 16:55
SamSam
212
$endgroup$
$begingroup$
I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
Mar 14 at 17:41
$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$
so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$
$$ =(m-nu)(m-nv) = m^2-mn+n^2$$
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
1
$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52
$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
Mar 14 at 19:25
$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
Mar 14 at 19:38
add a comment |
$begingroup$
Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$
so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$
$$ =(m-nu)(m-nv) = m^2-mn+n^2$$
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
1
$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52
$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
Mar 14 at 19:25
$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
Mar 14 at 19:38
add a comment |
$begingroup$
Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$
so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$
$$ =(m-nu)(m-nv) = m^2-mn+n^2$$
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
$endgroup$
Let $u$ and $v$ be zeros of $x^2-x+1=0$. Then $$x^2-x+1 = (x-u)(x-v)$$ and $$u^2=u-1;;;wedge ;;;;v^2=v-1$$
so $$(a^2 -ab + b^2)(c^2 - cd + d^2) = colorred(a-bu)colorblue(a-bv)colorred(c- du)colorblue(c-dv)$$
$$= colorredBig(ac+bdu^2-(ad+bc)uBig)colorblueBig(ac+bdv^2-(ad+bc)vBig)$$
$$= Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n uBig)Big(underbraceac-bd_m-underbrace(ad+bc-bd)_n vBig)$$
$$ =(m-nu)(m-nv) = m^2-mn+n^2$$
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
answered Mar 14 at 18:01
Maria MazurMaria Mazur
47.9k1260120
47.9k1260120
1
$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52
$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
Mar 14 at 19:25
$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
Mar 14 at 19:38
add a comment |
1
$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52
$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
Mar 14 at 19:25
$begingroup$
I'm sorry I'm not sure if they are relatively prime.
$endgroup$
– Maria Mazur
Mar 14 at 19:38
1
1
$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52
$begingroup$
Very nice how you bring in $u,vin mathbb C$ for manipulation purposes, and they drop right out at the end.
$endgroup$
– Keith Backman
Mar 14 at 18:52
$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
Mar 14 at 19:25
$begingroup$
Very nice indeed, thank you! Is it obvious that $m$ and $n$ are still coprime? Also, what happens if both $a^2 - ab + b^2$ and $c^2 - cd + d^2$ are divisible by $3$? Something should go wrong, because numbers of the form $m^2 - mn + n^2$ are never divisible by $9$ if $m$ and $n$ are coprime, right?
$endgroup$
– Al Tac
Mar 14 at 19:25
$begingroup$
I'm sorry I'm not sure if they are relatively prime.
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– Maria Mazur
Mar 14 at 19:38
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I'm sorry I'm not sure if they are relatively prime.
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– Maria Mazur
Mar 14 at 19:38
add a comment |
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There is this Identity:
$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$
Hence for:
$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$
$h=(ac+bd)$
$k=(bc-ad)$
$hk=(ac+bd)(bc-ad)$
Condition (c,d)=(2b,b-2a)
For $(a,b,c,d)=(3,7,14,1)$ we get:
$(49^2-49*95+95^2)=(37)*(183)=6771$
answered Mar 14 at 16:55
SamSam
212
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I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
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– Al Tac
Mar 14 at 17:41
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@Al Tac. There was a typo. See the edited answer above.
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– Sam
Mar 15 at 1:33
add a comment |
$begingroup$
There is this Identity:
$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$
Hence for:
$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$
$h=(ac+bd)$
$k=(bc-ad)$
$hk=(ac+bd)(bc-ad)$
Condition (c,d)=(2b,b-2a)
For $(a,b,c,d)=(3,7,14,1)$ we get:
$(49^2-49*95+95^2)=(37)*(183)=6771$
answered Mar 14 at 16:55
SamSam
212
$endgroup$
$begingroup$
I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
Mar 14 at 17:41
$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33
add a comment |
$begingroup$
There is this Identity:
$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$
Hence for:
$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$
$h=(ac+bd)$
$k=(bc-ad)$
$hk=(ac+bd)(bc-ad)$
Condition (c,d)=(2b,b-2a)
For $(a,b,c,d)=(3,7,14,1)$ we get:
$(49^2-49*95+95^2)=(37)*(183)=6771$
answered Mar 14 at 16:55
SamSam
212
$endgroup$
There is this Identity:
$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$
Hence for:
$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$
$h=(ac+bd)$
$k=(bc-ad)$
$hk=(ac+bd)(bc-ad)$
Condition (c,d)=(2b,b-2a)
For $(a,b,c,d)=(3,7,14,1)$ we get:
$(49^2-49*95+95^2)=(37)*(183)=6771$
answered Mar 14 at 16:55
SamSam
212
edited Mar 14 at 17:49
answered Mar 14 at 16:55
SamSam
212
answered Mar 14 at 16:55
SamSam
212
answered Mar 14 at 16:55
SamSam
212
212
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I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
Mar 14 at 17:41
$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33
add a comment |
$begingroup$
I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
Mar 14 at 17:41
$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33
$begingroup$
I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
Mar 14 at 17:41
$begingroup$
I am sorry, I am not sure I follow. What is your definition of $k$? If I take $h = ac + bd$ and $k = bc - ad$, in $h^2 - hk + k^2$ there is no term $abcd$, right? Also, in your example $k$ would be $-5$ not $+5$, so you would obtain $14^2 + 130 + 5^2$, it seems.
$endgroup$
– Al Tac
Mar 14 at 17:41
$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33
$begingroup$
@Al Tac. There was a typo. See the edited answer above.
$endgroup$
– Sam
Mar 15 at 1:33
add a comment |
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The prime divisors of such numbers are of the form $6n+1$, see here. So the product $(6n+1)(6m+1)$ is again of this form.
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– Dietrich Burde
Mar 14 at 16:20
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Yes but the product is not a prime anymore: are you saying that any number of the form $6x+1$ can be written as $h^2 - hk + k^2$? In the topic that you linked only the case of prime numbers is discussed, if I am not mistaken.
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– Al Tac
Mar 14 at 17:33
1
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Yes, it is not a prime anymore, but a product of primes of the form $6n+1$. It is represented by the quadratic form $x^2-xy+y^2$.
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– Dietrich Burde
Mar 14 at 17:56
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Thanks! I found a complete answer to the question of which numbers $n$ are represented as $h^2 - hk + k^2$ with $(h,k) = 1$ in Problem 3 of Section 3.7 of Niven-Zuckerman-Montgomery, An introduction to the theory of numbers. They are indeed the ones of the form $n = 3^alpha prod_i p_i^e_i$ with $alpha = 0, 1$ and $p_i equiv 1 pmod6$ for all $i$.
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– Al Tac
Mar 15 at 8:24