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I am trying to prove that not both integers $2^n-1,2^n+1$ can be prime for $n \not=2$. But I am not sure if my proof is correct or not:

Suppose both $2^n-1,2^n+1$ are prime, then $(2^n-1)(2^n+1)=4^n-1$ have 2 precisely two prim factors. Now $4^n-1=(4-1)(4^{n-1}+4^{n-2}+ \cdots +1)=3A$. So one of $2^n-1, 2^n+1$ must be 3. which implies $n=1$ or $n=2$ (rejected by assumption). Putting $n=1$, we have $2^n-1=1$ which is not a prime. Hence the result follows.

I also wanna know if there is alternative proof, thank you so much.

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    $\begingroup$ There is an alternative proof, $2^n - 1$ can only be prime if $n$ is prime, $2^n + 1$ can only be prime if $n$ is a power of $2$. And yes, your proof is correct. $\endgroup$ – Daniel Fischer Jul 25 '13 at 17:20
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    $\begingroup$ Just found : math.stackexchange.com/questions/402603/… $\endgroup$ – lab bhattacharjee Jul 25 '13 at 18:22
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    $\begingroup$ This is not a duplicate. This is a "check my proof" question, which cannot be a duplicate! $\endgroup$ – user1729 Jul 26 '13 at 9:49
  • $\begingroup$ (As a side point, your proof is fine.) $\endgroup$ – user1729 Jul 26 '13 at 9:50
  • $\begingroup$ @user1729: that's right. So, how can I recall my "close" vote? $\endgroup$ – mau Jul 26 '13 at 10:04
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Of the three consecutive integers $2^n-1,2^n,2^n+1$, one must be divisble by 3, and it can't be $2^n$.

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If $n$ is even $=2m,2^n-1=2^{2m}-1=4^m-1$ is divisible by $4-1=3$ and $4^m-1>3$ if $m\ge1\iff n\ge2$

If $n$ is odd $=2m+1,2^n+1=2^{2m+1}+1$ is divisible by $2+1=3$ and $2^{2m+1}+1>3$ if $m\ge1\iff n\ge3$


alternatively, $$(2^n-1)(2^n+1)=4^n-1$$ is divisible by $4-1=3$

So, at least one of $2^n-1,2^n+1$ is divisible by $3$

Now, $2^n+1>2^n-1>3$ for $n>2$

$\implies $ for $n>2,$ one of $2^n-1,2^n+1$ must be composite

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