Show that an integer of the form $8k + 7$ cannot be written as the sum of three squares.

Question

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 $$a^2 + b^2 + c^2 = 7 (mod 8)$$

Answer 1

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.

Answer 2

Note that $x^2\equiv (-x)^2\pmod 8$. So the squares mod $8$ are $0^2=0$, $1^2=1$, $2^2=4$ and $3^2=1$. It is evident that three of these numbers cannot add up $7$.

Answer 3

A variation:

Let $n = 8k + 7 = x^2 + y^2 +z^2$.

Then $n \equiv 3 \pmod 4$. Every square is congruent to $0$ or $1$ mod $4$, so $x^2, y^2, z^2 \equiv 1 \pmod 4$ for these numbers to add up to $ 3 \bmod 4$. Then $x, y, z$ are odd.

One checks immediately that $(\pm 3)^2 \equiv 1 \pmod 8, (\pm 1)^2 \equiv 1 \pmod 8$, which accounts for all odd numbers, so $n = x^2 + y^2 +z^2 \equiv 3 \pmod 8$. But this contradicts $n \equiv 7 \pmod 8$.

This is only a speedup up if you think it's faster to reduce mod $4$ to learn that you don't have to calculate $(\pm 2)^2 \bmod 8$ and $4^2 \bmod 4$.

Answer 4

Note that $$(2m+1)^2=8\frac {m(m+1)}2+1$$ and that $(2m)^2=4m^2$ is divisible by $4$