Prove that $n^7+7$ can never be a perfect square. Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n \equiv 5 \pmod{8}$ or $n \equiv 9 \pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.

probably easier than previous solutions:
Suppose that $n^7+7=x^2, x \in \mathbb{N}$ has a solution. Then we must have that $x^2+11^2=x^2+121=n^7+128=n^7+2^7$. Now $n+2|n^7+2^7=x^2+11^2$. If $11|n+2$, then $\nu_{11}(n^7+2^7)=\nu_{11}(n+2)$. This also means that $11|x^2+11^2 \implies 11|x$. Let $x=11y$, then we get $\nu_{11}(n+2)=\nu_{11}(n^7+2^7)=\nu_{11}(x^2+11^2)=2+\nu_{11}(y^2+1)=2$ since $-1$ is a quadratic non residue modulo $11$( because $11 \equiv 3 \pmod{4}$). If $11\not|n+2$ then obviously $\nu_{11}(n+2)=0$. in any case $\nu_{11}(n+2)$ is even. Consider any prime $p$ apart from $11$ dividing $n+2$, we must have that $x^2 \equiv -11^2 \pmod{p} \implies (x \cdot (11^{-1}))^2 \equiv -1 \pmod{p} \implies p \equiv 1 \pmod{4}$ because $-1$ is a quadratic residue  for only those primes which are $\equiv 1 \pmod{4}$. Hence we get $$n+2 \equiv 11^{2k}\equiv 1 \pmod{4} \implies n \equiv -1 \pmod{4}$$ where $k=0,1$. But then $x^2=n^7+7 \equiv 7-1 \equiv 2 \pmod{4}$ which is clearly false. Hence we are done.
