<s>[INST] Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares. I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed.
My attempt at the program was as follows: 
A number of the form, $8k + 7 = 7 (mod 8)$. That is, we are looking for integers a, b, c such that $a^2 + b^2 + c^2 = 7 (mod 8)$. 
LONG and TEDIOUS way:
$$(8k)^2 = 0 (mod 8)$$
$$(8k+1)^2 = 1 (mod 8)$$
$$(8k+2)^2 = 4 (mod 8)$$
$$(8k+3)^2 = 1 (mod 8)$$
$$(8k+4)^2 = 0 (mod 8)$$
$$(8k+5)^2 = 1 (mod 8)$$
$$(8k+6)^2 = 4 (mod 8)$$
$$(8k+7)^2 = 1 (mod 8)$$
That is, using three of these modulo there is no way to arrive at 
$$a^2 + b^2 + c^2 = 7 (mod 8)$$
 [/INST] You get a little bonus: $x^2 + y^2 + z^2$ can be even with two of the variables odd. However, $x^2 + y^2 + z^2$ cannot be divisible by $4$ unless all three of $x,y,z$ are even. So, Assume $x^2 + y^2 + z^2 \equiv 28 \pmod {32}.$ It follows that $x,y,z$ are even, and we get $(x/2)^2 + (y/2)^2 + (z/2)^2 \equiv 7 \pmod 8.$ This is a contradiction, so the sum of three squares cannot be $28 \pmod {32}.$ Do it again, the sum cannot be $112 \pmod {128}.$ And induction...the traditional way to write this is $$ x^2 + y^2 + z^2 \neq 4^k \cdot (8n+7).  $$ On the other hand, every other positive integer $m$ can be written as $m=x^2 + y^2 + z^2$ with integer variables. Gauss.
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