Integer as Sum of Three Squares

Theorem
Let $$r$$ be a positive integer.

Then $$r$$ can be expressed as the sum of three squares iff it is not of the form:
 * $$4^n \left({8 m + 7}\right)$$

for some $$m, n \in \Z: m \ge 0, n \ge 0$$.

Sufficient Condition
Suppose $$r$$ is not of the form $$4^n \left({8 m + 7}\right)$$.

Then we need to show that it can always be expressed as the sum of three squares.

Necessary Condition
From Square Modulo 8, the squares modulo $$8$$ are $$0, 1$$ and $$4$$.

So the sum of three squares can be congruent modulo 8 to any of the values $$0, 1, 2, 3, 4, 5$$ or $$6$$, but not $$7$$.

So no number of the form $$8 m + 7$$ can be the sum of three squares.

Now, suppose that $$\exists n \ge 1, m \ge 0$$ such that:
 * $$4^n \left({8 m + 7}\right) = x^2 + y^2 + z^2$$.

As the LHS is congruent modulo 4 to $$0$$, and as squares modulo 4 are either $$0$$ or $$1$$, it must be that $$x, y$$ and $$z$$ are all even.

Putting $$x = 2 x_1, y = 2 y_1, z = 2 z_1$$, we get:
 * $$4^{n-1} \left({8 m + 7}\right) = x_1^2 + y_1^2 + z_1^2$$

If $$n-1 > 1$$ then $$x_1, y_1$$ and $$z_1$$ are all still even, and the argument can be repeated:
 * $$4^{n-2} \left({8 m + 7}\right) = x_2^2 + y_2^2 + z_2^2$$

Thus we descend through all powers of $$4$$ till $$8 m + 7$$ itself is expressed as the sum of three squares.

But this is impossible, as we saw above.

So the assumption that $$4^n \left({8 m + 7}\right)$$ can be expressed as the sum of three squares is false.