(Number theory) Sum of three squares solution proof

[Xizel]

Homework Statement

Find all integer solutions to x² + y² + z² = 51. Use "without loss of generality."

Homework Equations

The Attempt at a Solution

My informal proof attempt:

Let x, y, z be some integers such that x, y, z = (0 or 1 or 2 or 3) mod 4
Then x², y², y² = (0 or 1) mod 4
So x² + y² + z² = [ (0 or 1) + (0 or 1) + (0 or 1) ] mod 4
Since 51 = 3 (mod 4) = x² + y² + z², then x, y, z = (1 or 3) mod 4

It is obvious that the solution is the permutations of $\pm1, \pm1, \pm7$.

It's my first proof course and I'm a little shaky. Is my logic correct? I feel like I took a leap from "Since 51..." to the solution, is there a more formal way to write that? I'm also not sure how to use wlog here. Thanks.

 

[fresh_42]

It is obvious that the solution is the permutations of $\pm1, \pm1, \pm7$.

I have a additional solution.

 

[Xizel]

I have a additional solution.

If my logic is correct, then x, y, z can take on values of 1, 3, 5, 7. I gave it some thought and I'm not seeing it

 

[fresh_42]

$2\cdot 25 + 1$
The remark "use w.l.o.g." probably refers to the assumption $x \geq y \geq z \geq 0$ which you could make.