If $2^n - 1$ is prime from some integer $n$, prove that n must also be prime.$2^p -1$ then $p$ is a primeIf $2^k - 1$ is prime then $k$ is prime, using $2^m -1 vert 2^mn - 1$$2^n + 1$ is prime $ implies n$ is a power of $2$Showing that $a = 2^1264504 - 1$ is not primeHow to prove that $n$ is a prime if $2^n-1$ is a primeIf $2^n-1$ is prime, then n is prime - proof involving the Mersenne primes by counterexampleIf an integer is divisible by 8 and 15, then the integer also must be divisible by which of the following?Prove that $1$ has only one divisorProve if $3$ does not divide $n$, then $n^2=1+3k$ for some integer $k$Prove for each positive integer $n$, there exists $n$ consecutive positive integers none of which is an integral power of a prime numberProving that no integer has multiplicative order 40 (mod 100)The integer $m$ is odd if and only if there exists $q in mathbbZ$ such that $m = 2q + 1$$A$ must contain two relatively prime elementsProve that for all prime numbers $ a,b,c , a^2 + b^2 neq c^2$Find all integers $x$ such that $frac2x^2-33x-1$ is an integer.Prove that integer not divisible by 2 or 3 is not divisible by 6
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A sequence that has integer values for prime indexes only:
If $2^n - 1$ is prime from some integer $n$, prove that n must also be prime.
$2^p -1$ then $p$ is a primeIf $2^k - 1$ is prime then $k$ is prime, using $2^m -1 vert 2^mn - 1$$2^n + 1$ is prime $ implies n$ is a power of $2$Showing that $a = 2^1264504 - 1$ is not primeHow to prove that $n$ is a prime if $2^n-1$ is a primeIf $2^n-1$ is prime, then n is prime - proof involving the Mersenne primes by counterexampleIf an integer is divisible by 8 and 15, then the integer also must be divisible by which of the following?Prove that $1$ has only one divisorProve if $3$ does not divide $n$, then $n^2=1+3k$ for some integer $k$Prove for each positive integer $n$, there exists $n$ consecutive positive integers none of which is an integral power of a prime numberProving that no integer has multiplicative order 40 (mod 100)The integer $m$ is odd if and only if there exists $q in mathbbZ$ such that $m = 2q + 1$$A$ must contain two relatively prime elementsProve that for all prime numbers $ a,b,c , a^2 + b^2 neq c^2$Find all integers $x$ such that $frac2x^2-33x-1$ is an integer.Prove that integer not divisible by 2 or 3 is not divisible by 6
$begingroup$
I understand the idea of the proof. I just want to make sure I wrote my proof well.
Suppose $n$ is not prime. Then $exists x,y in mathbbZ$ such that $n = xy$.
$2^xy - 1 = (2^x)^y - 1$
$ = (2^y - 1)(2^y(x-1) + 2^y(x-2) + ... + 2^y + 1)$
Since $2^n - 1$ is divisible by $2^y - 1$ it must be that $2^n - 1$ is not prime. Contradiction. Thus $n$ must be prime.
How does this look?
number-theory divisibility
asked Mar 3 '13 at 23:57
user64013user64013
219128
$endgroup$
|
show 3 more comments
$begingroup$
I understand the idea of the proof. I just want to make sure I wrote my proof well.
Suppose $n$ is not prime. Then $exists x,y in mathbbZ$ such that $n = xy$.
$2^xy - 1 = (2^x)^y - 1$
$ = (2^y - 1)(2^y(x-1) + 2^y(x-2) + ... + 2^y + 1)$
Since $2^n - 1$ is divisible by $2^y - 1$ it must be that $2^n - 1$ is not prime. Contradiction. Thus $n$ must be prime.
How does this look?
number-theory divisibility
asked Mar 3 '13 at 23:57
user64013user64013
219128
$endgroup$
6
$begingroup$
Exactly right. Similarly if $2^n+1$ is prime then $n$ is a power of two.
$endgroup$
– anon
Mar 4 '13 at 0:00
4
$begingroup$
Don't say in $mathbbZ$, that include $(-2)(-3)=6$. And it is positive integer. And in principle you must take care of the non-prime $n=1$. And you must insist that $x$ and $y$ are $gt 1$.
$endgroup$
– André Nicolas
Mar 4 '13 at 0:19
$begingroup$
Formally, you also need to say a word about $n=1$ (which does not satisfy the hypothesis, but since you are going by contrapositive and $n=1$ does satisfy the hypothesis of the contrapositive, namely $n$ is not prime, you need to dismiss the case (by saying $2^1-1$ is not prime) before introducing $x,y$).
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:41
$begingroup$
To make the proof really complete, you have to assume $x, y > 1$, so that $2^y - 1 > 1$, and $2^y - 1$ is a proper divisor of $2^n - 1$.
$endgroup$
– Andreas Caranti
Sep 17 '13 at 11:42
$begingroup$
@AndreasCaranti: and $2^y(x-1) + 2^y(x-2) + ... + 2^y + 1>1$ (since $x>1$) so the other factor is also a proper divisor.
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:44
|
show 3 more comments
$begingroup$
I understand the idea of the proof. I just want to make sure I wrote my proof well.
Suppose $n$ is not prime. Then $exists x,y in mathbbZ$ such that $n = xy$.
$2^xy - 1 = (2^x)^y - 1$
$ = (2^y - 1)(2^y(x-1) + 2^y(x-2) + ... + 2^y + 1)$
Since $2^n - 1$ is divisible by $2^y - 1$ it must be that $2^n - 1$ is not prime. Contradiction. Thus $n$ must be prime.
How does this look?
number-theory divisibility
asked Mar 3 '13 at 23:57
user64013user64013
219128
$endgroup$
I understand the idea of the proof. I just want to make sure I wrote my proof well.
Suppose $n$ is not prime. Then $exists x,y in mathbbZ$ such that $n = xy$.
$2^xy - 1 = (2^x)^y - 1$
$ = (2^y - 1)(2^y(x-1) + 2^y(x-2) + ... + 2^y + 1)$
Since $2^n - 1$ is divisible by $2^y - 1$ it must be that $2^n - 1$ is not prime. Contradiction. Thus $n$ must be prime.
How does this look?
number-theory divisibility
number-theory divisibility
asked Mar 3 '13 at 23:57
user64013user64013
219128
asked Mar 3 '13 at 23:57
user64013user64013
219128
asked Mar 3 '13 at 23:57
user64013user64013
219128
asked Mar 3 '13 at 23:57
user64013user64013
219128
asked Mar 3 '13 at 23:57
user64013user64013
219128
219128
6
$begingroup$
Exactly right. Similarly if $2^n+1$ is prime then $n$ is a power of two.
$endgroup$
– anon
Mar 4 '13 at 0:00
4
$begingroup$
Don't say in $mathbbZ$, that include $(-2)(-3)=6$. And it is positive integer. And in principle you must take care of the non-prime $n=1$. And you must insist that $x$ and $y$ are $gt 1$.
$endgroup$
– André Nicolas
Mar 4 '13 at 0:19
$begingroup$
Formally, you also need to say a word about $n=1$ (which does not satisfy the hypothesis, but since you are going by contrapositive and $n=1$ does satisfy the hypothesis of the contrapositive, namely $n$ is not prime, you need to dismiss the case (by saying $2^1-1$ is not prime) before introducing $x,y$).
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:41
$begingroup$
To make the proof really complete, you have to assume $x, y > 1$, so that $2^y - 1 > 1$, and $2^y - 1$ is a proper divisor of $2^n - 1$.
$endgroup$
– Andreas Caranti
Sep 17 '13 at 11:42
$begingroup$
@AndreasCaranti: and $2^y(x-1) + 2^y(x-2) + ... + 2^y + 1>1$ (since $x>1$) so the other factor is also a proper divisor.
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:44
|
show 3 more comments
6
$begingroup$
Exactly right. Similarly if $2^n+1$ is prime then $n$ is a power of two.
$endgroup$
– anon
Mar 4 '13 at 0:00
4
$begingroup$
Don't say in $mathbbZ$, that include $(-2)(-3)=6$. And it is positive integer. And in principle you must take care of the non-prime $n=1$. And you must insist that $x$ and $y$ are $gt 1$.
$endgroup$
– André Nicolas
Mar 4 '13 at 0:19
$begingroup$
Formally, you also need to say a word about $n=1$ (which does not satisfy the hypothesis, but since you are going by contrapositive and $n=1$ does satisfy the hypothesis of the contrapositive, namely $n$ is not prime, you need to dismiss the case (by saying $2^1-1$ is not prime) before introducing $x,y$).
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:41
$begingroup$
To make the proof really complete, you have to assume $x, y > 1$, so that $2^y - 1 > 1$, and $2^y - 1$ is a proper divisor of $2^n - 1$.
$endgroup$
– Andreas Caranti
Sep 17 '13 at 11:42
$begingroup$
@AndreasCaranti: and $2^y(x-1) + 2^y(x-2) + ... + 2^y + 1>1$ (since $x>1$) so the other factor is also a proper divisor.
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:44
6
6
$begingroup$
Exactly right. Similarly if $2^n+1$ is prime then $n$ is a power of two.
$endgroup$
– anon
Mar 4 '13 at 0:00
$begingroup$
Exactly right. Similarly if $2^n+1$ is prime then $n$ is a power of two.
$endgroup$
– anon
Mar 4 '13 at 0:00
4
4
$begingroup$
Don't say in $mathbbZ$, that include $(-2)(-3)=6$. And it is positive integer. And in principle you must take care of the non-prime $n=1$. And you must insist that $x$ and $y$ are $gt 1$.
$endgroup$
– André Nicolas
Mar 4 '13 at 0:19
$begingroup$
Don't say in $mathbbZ$, that include $(-2)(-3)=6$. And it is positive integer. And in principle you must take care of the non-prime $n=1$. And you must insist that $x$ and $y$ are $gt 1$.
$endgroup$
– André Nicolas
Mar 4 '13 at 0:19
$begingroup$
Formally, you also need to say a word about $n=1$ (which does not satisfy the hypothesis, but since you are going by contrapositive and $n=1$ does satisfy the hypothesis of the contrapositive, namely $n$ is not prime, you need to dismiss the case (by saying $2^1-1$ is not prime) before introducing $x,y$).
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:41
$begingroup$
Formally, you also need to say a word about $n=1$ (which does not satisfy the hypothesis, but since you are going by contrapositive and $n=1$ does satisfy the hypothesis of the contrapositive, namely $n$ is not prime, you need to dismiss the case (by saying $2^1-1$ is not prime) before introducing $x,y$).
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:41
$begingroup$
To make the proof really complete, you have to assume $x, y > 1$, so that $2^y - 1 > 1$, and $2^y - 1$ is a proper divisor of $2^n - 1$.
$endgroup$
– Andreas Caranti
Sep 17 '13 at 11:42
$begingroup$
To make the proof really complete, you have to assume $x, y > 1$, so that $2^y - 1 > 1$, and $2^y - 1$ is a proper divisor of $2^n - 1$.
$endgroup$
– Andreas Caranti
Sep 17 '13 at 11:42
$begingroup$
@AndreasCaranti: and $2^y(x-1) + 2^y(x-2) + ... + 2^y + 1>1$ (since $x>1$) so the other factor is also a proper divisor.
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:44
$begingroup$
@AndreasCaranti: and $2^y(x-1) + 2^y(x-2) + ... + 2^y + 1>1$ (since $x>1$) so the other factor is also a proper divisor.
$endgroup$
– Marc van Leeuwen
Sep 17 '13 at 11:44
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I'll try to recapituate all the comments made in an adapted version of the proof.
If $2^n−1$ is prime from some integer$~n$, then $n$ must also be prime.
Since the hypothesis requires $ngeq2$ to be true, one may assume that.
We prove the contrapositive: if $ngeq2$ is not prime then $2^n-1$ is not prime either.
Suppose $ngeq2$ is not prime. Then there exist integers $x,y>1$ such that $n = xy$.
One has $2^n-1 = 2^yx - 1 = (2^y)^x - 1 = (2^y - 1)(2^y(x-1) + 2^y(x-2) + cdots + 2^y.1 + 2^y.0)$. Since $y>1$ the first factor $2^y - 1$ is$>1$, and since $x>1$ that factor is less than $2^n-1$.
Since $2^n - 1$ has a proper divisor $2^y - 1$ greater than$~1$, we have shown that $2^n - 1$ is composite, establishing the contrapositive.
Remark. The contrapositive statement proved remains true if powers of $2$ are replaced throughout by powers of some integer $ageq2$. However with that change it is no longer true that the original hypothesis requires $ngeq2$, as $n=1$ might work. Therefore such a generalisation of the original statement requires an explicit additional hypothesis $ngeq2$.
$endgroup$
add a comment |
$begingroup$
Perhaps a minor nit-pick, but it seems the middle step should be $((2^y)^x-1)$, since you're utilizing the identity $$a^x-1=(a-1)(a^x-1+a^x-2+cdots+a+1)$$ with $a=2^y$.
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
This feels more like proof by contrapositive than proof by contradiction. Also, although obvious, without conditions on n,x,y it is not clear that $2^y-1$ can't be 1.
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
$endgroup$
$begingroup$
Should I say that $x$ and $y$ have to be greater then $1$ then?
$endgroup$
– user64013
Mar 4 '13 at 0:11
$begingroup$
yes, and you are also implicitly assuming $ngt1$
$endgroup$
– newToProgramming
Mar 4 '13 at 0:15
$begingroup$
Whats the reasoning x and y cant be 1?
$endgroup$
– user64013
Mar 4 '13 at 3:06
$begingroup$
In your proof, you state $2^n-1$ is divisible by $2^y-1$ and use this to conclude that $2^n-1$ is not prime, however consider what happens if $y$ is 1, can you still justify that same conclusion?
$endgroup$
– newToProgramming
Mar 4 '13 at 3:16
add a comment |
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3 Answers
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3 Answers
3
active
oldest
votes
active
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$begingroup$
I'll try to recapituate all the comments made in an adapted version of the proof.
If $2^n−1$ is prime from some integer$~n$, then $n$ must also be prime.
Since the hypothesis requires $ngeq2$ to be true, one may assume that.
We prove the contrapositive: if $ngeq2$ is not prime then $2^n-1$ is not prime either.
Suppose $ngeq2$ is not prime. Then there exist integers $x,y>1$ such that $n = xy$.
One has $2^n-1 = 2^yx - 1 = (2^y)^x - 1 = (2^y - 1)(2^y(x-1) + 2^y(x-2) + cdots + 2^y.1 + 2^y.0)$. Since $y>1$ the first factor $2^y - 1$ is$>1$, and since $x>1$ that factor is less than $2^n-1$.
Since $2^n - 1$ has a proper divisor $2^y - 1$ greater than$~1$, we have shown that $2^n - 1$ is composite, establishing the contrapositive.
Remark. The contrapositive statement proved remains true if powers of $2$ are replaced throughout by powers of some integer $ageq2$. However with that change it is no longer true that the original hypothesis requires $ngeq2$, as $n=1$ might work. Therefore such a generalisation of the original statement requires an explicit additional hypothesis $ngeq2$.
$endgroup$
add a comment |
$begingroup$
I'll try to recapituate all the comments made in an adapted version of the proof.
If $2^n−1$ is prime from some integer$~n$, then $n$ must also be prime.
Since the hypothesis requires $ngeq2$ to be true, one may assume that.
We prove the contrapositive: if $ngeq2$ is not prime then $2^n-1$ is not prime either.
Suppose $ngeq2$ is not prime. Then there exist integers $x,y>1$ such that $n = xy$.
One has $2^n-1 = 2^yx - 1 = (2^y)^x - 1 = (2^y - 1)(2^y(x-1) + 2^y(x-2) + cdots + 2^y.1 + 2^y.0)$. Since $y>1$ the first factor $2^y - 1$ is$>1$, and since $x>1$ that factor is less than $2^n-1$.
Since $2^n - 1$ has a proper divisor $2^y - 1$ greater than$~1$, we have shown that $2^n - 1$ is composite, establishing the contrapositive.
Remark. The contrapositive statement proved remains true if powers of $2$ are replaced throughout by powers of some integer $ageq2$. However with that change it is no longer true that the original hypothesis requires $ngeq2$, as $n=1$ might work. Therefore such a generalisation of the original statement requires an explicit additional hypothesis $ngeq2$.
$endgroup$
add a comment |
$begingroup$
I'll try to recapituate all the comments made in an adapted version of the proof.
If $2^n−1$ is prime from some integer$~n$, then $n$ must also be prime.
Since the hypothesis requires $ngeq2$ to be true, one may assume that.
We prove the contrapositive: if $ngeq2$ is not prime then $2^n-1$ is not prime either.
Suppose $ngeq2$ is not prime. Then there exist integers $x,y>1$ such that $n = xy$.
One has $2^n-1 = 2^yx - 1 = (2^y)^x - 1 = (2^y - 1)(2^y(x-1) + 2^y(x-2) + cdots + 2^y.1 + 2^y.0)$. Since $y>1$ the first factor $2^y - 1$ is$>1$, and since $x>1$ that factor is less than $2^n-1$.
Since $2^n - 1$ has a proper divisor $2^y - 1$ greater than$~1$, we have shown that $2^n - 1$ is composite, establishing the contrapositive.
Remark. The contrapositive statement proved remains true if powers of $2$ are replaced throughout by powers of some integer $ageq2$. However with that change it is no longer true that the original hypothesis requires $ngeq2$, as $n=1$ might work. Therefore such a generalisation of the original statement requires an explicit additional hypothesis $ngeq2$.
$endgroup$
I'll try to recapituate all the comments made in an adapted version of the proof.
If $2^n−1$ is prime from some integer$~n$, then $n$ must also be prime.
Since the hypothesis requires $ngeq2$ to be true, one may assume that.
We prove the contrapositive: if $ngeq2$ is not prime then $2^n-1$ is not prime either.
Suppose $ngeq2$ is not prime. Then there exist integers $x,y>1$ such that $n = xy$.
One has $2^n-1 = 2^yx - 1 = (2^y)^x - 1 = (2^y - 1)(2^y(x-1) + 2^y(x-2) + cdots + 2^y.1 + 2^y.0)$. Since $y>1$ the first factor $2^y - 1$ is$>1$, and since $x>1$ that factor is less than $2^n-1$.
Since $2^n - 1$ has a proper divisor $2^y - 1$ greater than$~1$, we have shown that $2^n - 1$ is composite, establishing the contrapositive.
Remark. The contrapositive statement proved remains true if powers of $2$ are replaced throughout by powers of some integer $ageq2$. However with that change it is no longer true that the original hypothesis requires $ngeq2$, as $n=1$ might work. Therefore such a generalisation of the original statement requires an explicit additional hypothesis $ngeq2$.
edited Sep 18 '13 at 13:37
community wiki
3 revs
Marc van Leeuwen
add a comment |
add a comment |
$begingroup$
Perhaps a minor nit-pick, but it seems the middle step should be $((2^y)^x-1)$, since you're utilizing the identity $$a^x-1=(a-1)(a^x-1+a^x-2+cdots+a+1)$$ with $a=2^y$.
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
Perhaps a minor nit-pick, but it seems the middle step should be $((2^y)^x-1)$, since you're utilizing the identity $$a^x-1=(a-1)(a^x-1+a^x-2+cdots+a+1)$$ with $a=2^y$.
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
add a comment |
$begingroup$
Perhaps a minor nit-pick, but it seems the middle step should be $((2^y)^x-1)$, since you're utilizing the identity $$a^x-1=(a-1)(a^x-1+a^x-2+cdots+a+1)$$ with $a=2^y$.
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
$endgroup$
Perhaps a minor nit-pick, but it seems the middle step should be $((2^y)^x-1)$, since you're utilizing the identity $$a^x-1=(a-1)(a^x-1+a^x-2+cdots+a+1)$$ with $a=2^y$.
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
answered Sep 17 '13 at 11:29
Rebecca J. StonesRebecca J. Stones
21k22781
21k22781
add a comment |
add a comment |
$begingroup$
This feels more like proof by contrapositive than proof by contradiction. Also, although obvious, without conditions on n,x,y it is not clear that $2^y-1$ can't be 1.
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
$endgroup$
$begingroup$
Should I say that $x$ and $y$ have to be greater then $1$ then?
$endgroup$
– user64013
Mar 4 '13 at 0:11
$begingroup$
yes, and you are also implicitly assuming $ngt1$
$endgroup$
– newToProgramming
Mar 4 '13 at 0:15
$begingroup$
Whats the reasoning x and y cant be 1?
$endgroup$
– user64013
Mar 4 '13 at 3:06
$begingroup$
In your proof, you state $2^n-1$ is divisible by $2^y-1$ and use this to conclude that $2^n-1$ is not prime, however consider what happens if $y$ is 1, can you still justify that same conclusion?
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– newToProgramming
Mar 4 '13 at 3:16
add a comment |
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This feels more like proof by contrapositive than proof by contradiction. Also, although obvious, without conditions on n,x,y it is not clear that $2^y-1$ can't be 1.
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
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Should I say that $x$ and $y$ have to be greater then $1$ then?
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– user64013
Mar 4 '13 at 0:11
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yes, and you are also implicitly assuming $ngt1$
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– newToProgramming
Mar 4 '13 at 0:15
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Whats the reasoning x and y cant be 1?
$endgroup$
– user64013
Mar 4 '13 at 3:06
$begingroup$
In your proof, you state $2^n-1$ is divisible by $2^y-1$ and use this to conclude that $2^n-1$ is not prime, however consider what happens if $y$ is 1, can you still justify that same conclusion?
$endgroup$
– newToProgramming
Mar 4 '13 at 3:16
add a comment |
$begingroup$
This feels more like proof by contrapositive than proof by contradiction. Also, although obvious, without conditions on n,x,y it is not clear that $2^y-1$ can't be 1.
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
$endgroup$
This feels more like proof by contrapositive than proof by contradiction. Also, although obvious, without conditions on n,x,y it is not clear that $2^y-1$ can't be 1.
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
answered Mar 4 '13 at 0:09
newToProgrammingnewToProgramming
1755
1755
$begingroup$
Should I say that $x$ and $y$ have to be greater then $1$ then?
$endgroup$
– user64013
Mar 4 '13 at 0:11
$begingroup$
yes, and you are also implicitly assuming $ngt1$
$endgroup$
– newToProgramming
Mar 4 '13 at 0:15
$begingroup$
Whats the reasoning x and y cant be 1?
$endgroup$
– user64013
Mar 4 '13 at 3:06
$begingroup$
In your proof, you state $2^n-1$ is divisible by $2^y-1$ and use this to conclude that $2^n-1$ is not prime, however consider what happens if $y$ is 1, can you still justify that same conclusion?
$endgroup$
– newToProgramming
Mar 4 '13 at 3:16
add a comment |
$begingroup$
Should I say that $x$ and $y$ have to be greater then $1$ then?
$endgroup$
– user64013
Mar 4 '13 at 0:11
$begingroup$
yes, and you are also implicitly assuming $ngt1$
$endgroup$
– newToProgramming
Mar 4 '13 at 0:15
$begingroup$
Whats the reasoning x and y cant be 1?
$endgroup$
– user64013
Mar 4 '13 at 3:06
$begingroup$
In your proof, you state $2^n-1$ is divisible by $2^y-1$ and use this to conclude that $2^n-1$ is not prime, however consider what happens if $y$ is 1, can you still justify that same conclusion?
$endgroup$
– newToProgramming
Mar 4 '13 at 3:16
$begingroup$
Should I say that $x$ and $y$ have to be greater then $1$ then?
$endgroup$
– user64013
Mar 4 '13 at 0:11
$begingroup$
Should I say that $x$ and $y$ have to be greater then $1$ then?
$endgroup$
– user64013
Mar 4 '13 at 0:11
$begingroup$
yes, and you are also implicitly assuming $ngt1$
$endgroup$
– newToProgramming
Mar 4 '13 at 0:15
$begingroup$
yes, and you are also implicitly assuming $ngt1$
$endgroup$
– newToProgramming
Mar 4 '13 at 0:15
$begingroup$
Whats the reasoning x and y cant be 1?
$endgroup$
– user64013
Mar 4 '13 at 3:06
$begingroup$
Whats the reasoning x and y cant be 1?
$endgroup$
– user64013
Mar 4 '13 at 3:06
$begingroup$
In your proof, you state $2^n-1$ is divisible by $2^y-1$ and use this to conclude that $2^n-1$ is not prime, however consider what happens if $y$ is 1, can you still justify that same conclusion?
$endgroup$
– newToProgramming
Mar 4 '13 at 3:16
$begingroup$
In your proof, you state $2^n-1$ is divisible by $2^y-1$ and use this to conclude that $2^n-1$ is not prime, however consider what happens if $y$ is 1, can you still justify that same conclusion?
$endgroup$
– newToProgramming
Mar 4 '13 at 3:16
add a comment |
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Exactly right. Similarly if $2^n+1$ is prime then $n$ is a power of two.
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– anon
Mar 4 '13 at 0:00
4
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Don't say in $mathbbZ$, that include $(-2)(-3)=6$. And it is positive integer. And in principle you must take care of the non-prime $n=1$. And you must insist that $x$ and $y$ are $gt 1$.
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– André Nicolas
Mar 4 '13 at 0:19
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Formally, you also need to say a word about $n=1$ (which does not satisfy the hypothesis, but since you are going by contrapositive and $n=1$ does satisfy the hypothesis of the contrapositive, namely $n$ is not prime, you need to dismiss the case (by saying $2^1-1$ is not prime) before introducing $x,y$).
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– Marc van Leeuwen
Sep 17 '13 at 11:41
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To make the proof really complete, you have to assume $x, y > 1$, so that $2^y - 1 > 1$, and $2^y - 1$ is a proper divisor of $2^n - 1$.
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– Andreas Caranti
Sep 17 '13 at 11:42
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@AndreasCaranti: and $2^y(x-1) + 2^y(x-2) + ... + 2^y + 1>1$ (since $x>1$) so the other factor is also a proper divisor.
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– Marc van Leeuwen
Sep 17 '13 at 11:44