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The actual description of the question is:
Suppose that $2^n + 1$ is a prime where $n$ is a positive integer. Prove that $n = 2^k$ for some positive integer $k$.

There is only one approach that know of:
To show that there is factorization in which at least one factor $(x-a)=1$, and the others are prime. A nice simplification would be to show that all except one root is having value such that the factor has value = $1$.
For example, if given quadratic: $x^2 + 2x - 3 = (x+3)(x-1)$. As have two factors, so one of these factors to be $1$, and the other prime.
This leads to:
$$x-1=1 \implies x=2 \implies (x+3)(x-1) = 5$$ else, $$x+3=1 \implies x = -2 \implies (x-1)(x+3) => -3$$

But, if consider positive integers then $x=2$ is the value.

if given quadratic: $x^2 - 5x +6 = (x-3)(x-2)$. As have two factors, so one of these factors to be $1$, and the other prime.
This leads to:
$$x-3=1 \implies x=4 \implies (x-3)(x-2) = 2$$ else, $$x-2=1 \implies x = 3 \implies (x-3)(x-2) => 0$$

But, if consider that $0$ is not a prime, then $x=4$ is the value.

So, let us try to work out cases with value of $k$ starting form $0$.
$k=0, 2^0=1, 2^1+1 =3$
$k=1, 2^1=2, 2^2 +1=5$

So, weak induction will suffice here, as strong induction is not needed as no need for previous values to prove new case.

So, let the assumption be that it holds for $k=l$, need prove for $k=l+1$,
but, $2^{2^{l+1}} = 2^{{2^l}^2} = (2^{2^l})^2$
So, if $2^{2^l} + 1$ is a prime, let p; then need prove $(p-1)^2+1$ is also.

i.e., need prove $p^2-2p+2$ is a prime, given $p$ is.

jitender
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    For example $7^2 - 7\cdot 5 +2=16$ – rtybase Jan 28 '18 at 12:00
  • @rtybase So, it is a failure. – jitender Jan 28 '18 at 12:01
  • Let's put it in a positive light, you just need another approach :) – rtybase Jan 28 '18 at 12:03
  • @rtybase Can the reason can be given algebraically without any value of $p$, as there is negative discriminant of $p^2 -2p+2 $. so, no real root is possible. Any other approach is not coming to me, can you please suggest. – jitender Jan 28 '18 at 12:07
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    Maybe ... but if you try to prove a statement which has counterexamples, you don't really have to look for an algebraic reason for that (unless you insist of course), the statement is wrong. I meant, you need a different strategy to prove that if $2^n+1$ is prime, then $n=2^k$ for some $k$. I find it easy to use modular arithmetics, plus learning modular arithmetics is not so difficult. – rtybase Jan 28 '18 at 12:12

2 Answers2

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We have:

$$2^{ab} + 1 = (2^{a} + 1)(2^{a(b-1)} - 2^{a(b-2)} + \cdots + 1)$$

where $b$ is an odd integer. It's obvious that both of the factors are bigger than 1 and so if $n$ has an odd factor we $2^n + 1$ has a nontrivial factor.

If you're more interested in primes of that form, they are called Fermat's Primes and till now only five of them are known.

Stefan4024
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  • in my edited OP, have got $p^2 -2p +2$ to be proved as a prime, if $p$ is. Please link your answer to that, if I am correct. – jitender Jan 28 '18 at 11:56
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Also from $2 \equiv -1 \pmod{3} \Rightarrow 2^{2k+1} \equiv (-1)^{2k+1} \equiv -1 \pmod{3}$. Since we ask for $2^n+1$ to be prime (thus it can't be divisible by 3, unless $n=1=2^0$), $n$ must be even or $n=2k_0$, then it becomes $4^{k_0}+1$ which is $4^{k_0} \equiv -1 \pmod{5}$ for odd $k_0$ so (unless $n=2^1$) it must be even $k_0=2k_1$ and $n=2^2k_1$ and so on by induction. This process stops for some $k_t$ s.t. $\frac{n}{2^{k_t}}\leq 1$.

rtybase
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