If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$. If $n = a^2 + b^2 + c^2$ for positive integers $a$, $b$,$c$, show that there exist positive integers $x$, $y$, $z$ such that $n^2 = x^2 + y^2 + z^2$.


I feel that the problem basically uses algebraic manipulation even though it's in a Number Theory textbook. I don't realize how to show $(a^2+b^2+c^2)^2$ as the sum of three squares. I have tried algebraic manipulation but this is the stage I have reached.

$$(b^2 + c^2)^2 + a^2(a^2 + b^2 + c^2 + b^2 + c^2)$$

Could you give me some hints on how to proceed with this question?

Thanks.
If:

$$x=2ac$$

$$y=2bc$$

$$z=a^2+b^2-c^2$$

$$n=a^2+b^2+c^2$$

Then:

$$x^2+y^2+z^2=n^2$$
A theoretical and easy proof for $n$ odd.

According to a theorem of Gauss the positive integer $n$ is a sum of three squares if and only if $n$ is not of the form $4^a(8b-1)$; $a,b\in \mathbb Z$.

If $n^2=8B-1$ then $n^2\equiv -1 \pmod8$. This is not possible because $-1$ is not a square modulo $8$.

Thus the odd integer $n^2$ is always a sum of three squares, in particular when $n$ is itself a sum of three squares.

It should not be very hard the proof for n even, but I do not try because the constructive and nice proof of @individ above.