ONTC Final Exam

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Masum
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Re: ONTC Final Exam

Unread post by Masum » Sun Sep 13, 2015 11:16 pm

Try this: $n^2-1|2^n+1$
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tanmoy
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Re: ONTC Final Exam

Unread post by tanmoy » Mon Sep 14, 2015 2:48 pm

$2^{n}+1$ is odd.So,$n^{2}-1$ is odd.So,$n$ is even.Let $n=2k$.Then $2^{n}+1=(2^{k})^{2}+1$.So,every divisor of $2^{n}+1$ is of the form $4m+1$.
So,$n^{2}-1 \equiv 1 (mod 4)$.
Or,$n^{2} \equiv 2 (mod 4)$,which is impossible.
So there is no solution. :)
"Questions we can't answer are far better than answers we can't question"

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Masum
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Re: ONTC Final Exam

Unread post by Masum » Mon Sep 14, 2015 3:26 pm

tanmoy wrote:$2^{n}+1$ is odd.So,$n^{2}-1$ is odd.So,$n$ is even.Let $n=2k$.Then $2^{n}+1=(2^{k})^{2}+1$.So,every divisor of $2^{n}+1$ is of the form $4m+1$.
So,$n^{2}-1 \equiv 1 (mod 4)$.
Or,$n^{2} \equiv 2 (mod 4)$,which is impossible.
So there is no solution. :)
You have to prove it.
One one thing is neutral in the universe, that is $0$.

tanmoy
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Re: ONTC Final Exam

Unread post by tanmoy » Mon Sep 14, 2015 7:19 pm

Okay.Suppose,$p$ is an odd divisor of $2^{n}+1$.So, $p$ is either of the form $4m+1$ or $4m+3$,where $m$ any integer.Let $p$ is of the form $4m+3$.Then $2^{n} \equiv -1 (mod p)$.Or $2^{p-1} \equiv (2^{n})^{2m+1} \equiv -1 (mod p)$,which contradicts Fermat's little theorem.
So,$p$ is of the form $4m+1$.
Now,if $n^{2}-1$ is prime,we are done! But if $n^{2}-1$ is composite,then it has two or more odd prime divisors which have the form $4m+1$.So,there product is also of the form।But $n^{2}-1$ is of the form $4m+3$।So,contradiction.Is it correct now?
"Questions we can't answer are far better than answers we can't question"

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Masum
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Re: ONTC Final Exam

Unread post by Masum » Mon Sep 14, 2015 11:52 pm

now it should be correct
One one thing is neutral in the universe, that is $0$.

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