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I am trying to prove this by contradiction:

Assume $2^{k} + 1$ is prime.

By definition of odd number $2^{k} + 1$ is odd because $2^{k} + 1 = 2\cdot 2^{k-1} + 1$

Then $2^{k} + 1 \pmod{2} \equiv 1 \pmod{2}$

Because $2^{k} + 1$ is odd, $2^{k}$ has to be even. Let $k = ab$

I am stuck here, can someone help me?

Jyrki Lahtonen
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JOX
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    $a+1 | a^n+1$ when $n$ is odd. – r9m Jan 30 '15 at 15:38
  • How do I obtain that conclusion? I'm trying to manipulate the expression but I'm failing – JOX Jan 30 '15 at 15:57
  • @JOX take a look at my solution – Shobhit Jan 30 '15 at 15:58
  • @JOX $a^n+1 = (a+1)(a^{n-1}-a^{n-2}+\cdots + 1)$ when $n$ is odd (how? expand the RHS or identify it as a Goemetric progression with common ratio $(-a)$ .. do you know how to sum a GP?) – r9m Jan 30 '15 at 16:02

1 Answers1

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Hint:

If $k$ is odd then, $2^k+1=2^k-(-1)^k$, now use that $x^n-y^n $is divisible by $x-y$ to conclude.

Proof that $x^n-y^n$ is divisible by $x-y$:

We prove this by induction, $n=1$ follows, let $x^k-y^k$ be divisible by $x-y$.

Then $x^{k+1}-y^{k+1}=x(x^k)-x(y^k)+x(y^k)-y(y^k)=x(x^k-y^k)+y^k(x-y)$, which is divisible by $x-y$.

Hence by induction the result follows.

Shobhit
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  • I'm trying to understand your last part, can I obtain that divisibility by manipulating the expression? I don't seem to remember that fact about x-y from before – JOX Jan 30 '15 at 16:02
  • @JOX look at the edited answer – Shobhit Jan 30 '15 at 16:18