Look at the equation modulo 3 firstly: $(-1)^n + n^2 \equiv 0$ so $n\equiv 1$ or $2 \mod 3$. Then look modulo 7: $n^2 \equiv -2^n \equiv -2 $ or $-4$ (as $2^3\equiv 1 \mod 7$) $\equiv 5$ or $3$. But $5$ nor $3$ are quadratic residues modulo 7 hence no solutions. Is this problem really correct?

I wonder, too. This looks too easy... Omid Hatami, please tell us?

I think you forget that some of the variables can equal zero. Anyway, the problem is easy, but there are some missed cases, since I see that $n1$ works perfectly for instance.

Gentlemen(i mean the moderators , harazi ,megus & Arne ), I think the problem is in this form: $p,q \in \mathbb{N} \cup \{0\}$ As megus has showd one of the $p,q$ must be zero so we shall solve it in these forums: 1.$n^2=3^p-2^n$ 2.$n^2=7^q-2^n$

lomos_lupin wrote: Gentlemen(i mean the moderators , harazi ,megus & Arne ), I think the problem is in this form: $p,q \in \mathbb{N} \cup \{0\}$ As megus has showd one of the $p,q$ must be zero so we shall solve it in these forums: 1.$n^2=3^p-2^n$ 2.$n^2=7^q-2^n$ I think we can only consider $n^2=3^p-2^n$,since I think $n^2+2^n$ can't be divide by 7;I think p will be an odd number also. (I don't know is my thought is correct or not )

easy to prove that $q=0$.I have $n^{2}+2^{n}=3^{p}\Rightarrow p is even$ $< = > n^{2}-3^{2k}=2^{n}$ $< = >(n-3^{k})(n+3^{k})=2^{n}$.to here then simply. _________________________ $VietNam is always in my heart$

cuopbientcd wrote: easy to prove that $q=0$.I have $n^{2}+2^{n}=3^{p}\Rightarrow p is even$ $VietNam is always in my heart$ or p=1. All solutions $n=1$ and $n=3$.