# Help with a proof.

1. Jan 22, 2004

### gimpy

Let $$n \geq 2$$ be an integer such that $$2^n + n^2$$ is prime. Prove that $$n \equiv 3(mod 6)$$.

Ok this is what i have so far.
$$n \equiv 3(mod 6)\\ \Rightarrow 6|n-3\\ \Rightarrow n - 3 = 6k \\ \Rightarrow n = 6k + 3 \\$$.

Ok well i think i can use a proof by contradition. Obviosly $$n$$ cannot be an even number because $$2^n + n^2$$ wont be prime. So $$n$$ must be $$6k + 1$$ or $$6k + 3$$ or $$6k + 5$$. Now when i plug all these into $$2^n + n^2$$ and mess around with it for a bit i just can't seem to prove it. Can anyone help me out?

2. Jan 23, 2004

### matt grime

It suffices to consider the 2^n + n^2 mod 3.

Clearly, if n is NOT congruent to 0 mod three, then as n must be odd, 2^n = -1 mod 3 and n^2 = 1 mod 3, thus 2^n + n^2 =0 mod 3 and can't be prime (n=1 excepted)

Hence n is divisble by 3, and not by two, thus it is congruent to 3 mod 6

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