When $2^n-1$ is prime and $n>2$ then $n$ is prime. Then, when $2^n-1$ is prime, why $2^n+1$ is composite?
What I have done is this.
Let's suppose $2^n+1$ is prime, then it will be contradiction.
But I can't proceed any more
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2I think you've written something wrong. Look at $n = 2$; in this case, $2^n - 1$ is prime (3), but $2^n + 1$ is also prime (5). – John Hughes Jun 04 '16 at 10:06
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If $n$ is odd then $2^n+1$ is divisible by $3$. – lulu Jun 04 '16 at 10:07
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Using http://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd, $$2^n+1$$ will be divisible by $2+1$ if $n$ is odd – lab bhattacharjee Jun 04 '16 at 10:08
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1another condition is that n>2 – nien Jun 04 '16 at 10:09
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$$x^{2m+1}+1=(x+1)(x^{2m}-x^{2m-1}+x^{2m-2}-\dots-x+1).$$ So $2^n+1$ is divisible by $3$ id $n$ is odd (whether prime or not).
Of course this also can be found with congruences modulo $3$ , if you note $2\equiv -1\mod3$.
Bernard
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It is not true. A counterexample is $n=2$, when both of $2^n-1=3$ and $2^n+1=5$ are prime.
That is the only counterexample, though -- it requires a Fermat prime and a Mersenne prime with the same exponent, and the exponent in a Fermat prime is always a power of 2, whereas that for a Mersenne prime is always prime.
hmakholm left over Monica
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