Primes, congruence mod4, odd exponents, sum of two squares I am trying to prove:
(T) If a prime $p$ is congruent to $3 \bmod 4$ and it occurs with an odd exponent in the prime factorization of $n\in\mathbb N$, then $n$ is not a sum of two squares.
I have tried this idea (not sure it is going to be useful).
Let $p = 4k + 3$, where $k$ is a positive integer. So 
$p^{2m + 1} = (4k = 3)^{2m + 1} = \overbrace{(4k + 3)^2\dots(4k + 3)^2}^m(4k + 3)$
$ = \overbrace{(4t + 1)\dots(4t + 1)}^m(4k + 3)$ for some $t\in\mathbb N$
$ = (4r + 1)(4k + 3)$ for some $r\in\mathbb N$
$\equiv3 \bmod 4$
I know how to prove that a sum of two squares is never equivalent to $3 \bmod 4$. So, using this fact, $p^{2m + 1}$ is not the sum of two squares. 
Let $$n = \prod_{i = 1}^m {p_i}^{t_i}$$ where $p_i$ are primes and $t_i$ positive integers. Let $p = 4k + 3 = p_i$ and $t_1 = 2m + 1$. So $$n = (4r + 1)(4k + 3) \prod_{j = 1}^{i - 1} {p_j}^{t_j} \prod_{j = i + 1}^m {p_j}^{t_j}$$.
If I can prove that $n \equiv 3 \bmod 4$ I am done. However, as I said, I am not sure that this idea is going to be useful.
 A: Here is a number theory approach my teacher showed me
Let the prime factorization of $n$ be
$$n=2^\alpha p_1^{\beta_1}p_2^{\beta_2}...p_r^{\beta_r}q_1^{\gamma_1}q_2^{\gamma_2}...q_s^{\gamma_s}$$
such that $p_i\equiv1 \pmod 4$ , ($1\le i \le r$) and $q_j \equiv 3 \pmod 4$ ($1\le j\le s$). Now take at least one of the $\gamma_i$ to be odd, we'll say $\gamma_1$. If $n=x^2+y^2$ and $d= \gcd(x,y)$, then with $n_1=\frac{n}{d^2}$, $x_0=\frac xd$, and $y_0 =\frac yd$ we know 
$$n_1=\frac n{d^2}=\left( \frac xd \right)^2 +\left( \frac yd \right)^2=x_0^2+y_0^2 $$
now let $\tilde \gamma$ be the exponent of $q_1$ in the prime factorization of $n_1$. $\tilde \gamma$ is odd and $\gcd(x_0,q_1)=\gcd(y_0,q_1)=1$ so there exists an integer $a$ such that 
$$x_0 \equiv y_0a \pmod{q_1} \\\implies 0\equiv n_1=x_0^2+y_0^2\equiv a^2y_0^2+y_0^2\equiv y_0^2(1+a^2) \pmod{q_1}\tag1$$
$q_1$ does not divide $ y_0^2$ so 
$$a^2+1\equiv0 \pmod{q_1} \implies a^2 \equiv -1 \pmod{q_1}$$
which cannot have a solution by Gauss' Lemma. Contradiction.

$(1)$ because if $d=\gcd(a,c)$ then the congruence $b\equiv an \pmod c$ has $d$ mutually incongruent solutions if $d \mid b$
