Proofs for couple of details when proving that $n^7 + 7$ is not a perfect square for any integer $n$. 
Prove that $n^7 + 7$ is not a perfect square for any integer $n$.

The proof given for the problem is the following

Assume $n > 0$ since $n = 0$ and $n = -1$ are easy and for $n \le -2$ the expression is negative. Suppose $n^7 + 7 = a^2$. Then $$n^7+2^7 = a^2+11^2.$$ Taking modulo $4$ gives $n ≡ 1 \pmod 4$, but $n + 2 \mid a^2 + 11^2$, and $n + 2 ≡ 3 \pmod4$. Note that $a^2 + 11^2$ has no $3$ mod $4$ prime factors except possibly an $11^2$, by Fermat’s Christmas theorem. Since $n + 2 ≡ 3 \pmod 4$ we would need to have $\nu_{11}(n + 2) = 1$ as a result, since $\nu_{11}(n + 2)$ should be odd and at most $2$. However, we then get $$\nu_{11}(a^2+11^2)=\nu_{11}(n^7+2^7)=\nu_{11}(n+2)+\nu_{11}(7)=\nu_{11}(n+2)=1$$
by the exponent lifting lemma, which is impossible.

There are few parts I don't understand. The first one is the line

Note that $a^2 + 11^2$ has no $3$ mod $4$ prime factors except possibly an $11^2$, by Fermat’s Christmas theorem.

I know that we have since $n+2 \mid a^2+11^2$ and $n+2 \equiv 3 \pmod{4}$ the term $n+2$ must have a prime divisor $p$ such that $p \equiv 3 \pmod 4$ and now by Fermat’s Christmas theorem $p \mid n+2 \implies p \mid a^2 +11^2 \implies p \mid a$ and $p \mid 11$ implying that $p = 11$. But why is this stating that it could have $11^2$ as a factor and no other factors which are $3$ mod $4$?
The second fact I don't understand is that

Since $n + 2 ≡ 3 \pmod 4$ we would need to have $\nu_{11}(n + 2) = 1$ as a result, since $\nu_{11}(n + 2)$ should be odd and at most $2$.

Why should $\nu_{11}(n+2)=1$ and what makes $\nu_{11}(n+2)$ odd and at most $2$?
 A: We first show that $\nu_{11}(a^2+11^2)$, and thus, $\nu_{11}(n+2)$,  is at most $2$. Now, on the one hand, for $\nu_{11}(a^2+11^2)$ to be positive at least $2$ in the first place, $a^2$ and thus $a$ itself must also be a multiple of $11$, say $a=11c$ for some integer $c$. So $a^2+11^2$ can be written $a^2+11^2 = (11c)^2+11^2 = 11^2(c^2+1)$. However, on the other hand, for $11^3$ to divide $11^2(c^2+1)$, it follows that the integer $c$ must must be such $11$ that divides $c^2+1$, which gives $c^2 \equiv_{11} -1$, or equivalently, $-1$ a square in $\mathbb{Z}/11\mathbb{Z}$, which is impossible, as $11 \pmod 4 = 3$. So $11$ cannot divide $c^2+1$ after all, and thus $11^3$ cannot divide $11^2(c^2+1)=a^2+11^2$ after all, and thus indeed, $\nu_{11}(a^2+11^2)$ is at most $2$.
Finally, $\nu_{11}(n+2)$ can be at most $2$ as well, as  $\nu_{11}(a^2+11^2)$ is at most $2$ and $n+2$ divides $a^2+11$.
To see that $11$ is the only prime $p$ satisfying $p \pmod 4 =3$ dividing $a^2+11^2$, note that $p|(a^2+11^2)$ gives $a^2 \equiv_{p} -(11^2),$ which implies either that

*

*Either $-1$ is a square in $\mathbb{Z}/p\mathbb{Z}$ [impossible for all primes $p$ such that $p \pmod 4 = 3$];


*Or $11^2 \equiv_p 0$ which gives $11=p$.
We now show that $\nu_{11}(n+2)$ must be odd.  Now, as shown above, $11$ is the only prime $p$ satisfying $p \pmod 4 =3$ that may divide $n+2$. As $n+2$ is odd, it follows that every other prime $q$ dividing $n+2$ satisfies $q \pmod 4 = 1$, and thus, $$n+2 \pmod 4 = 11^{\nu_{11}(n+2)} \pmod 4.$$
As $n+2 \pmod 4$ is $3$ and $11^e \pmod 4$ is $1$ for all even nonnegative integers $e$, it follows that $\nu_{11}(n+2)$ must indeed be odd.
