Solving for $a,b,c,d$ where $a^2 + b^2 + c^2 + d^2 = 630^2$ How could one solve for $a,b,c,d$ where:
$$a^2 + b^2 + c^2 + d^2 = 630^2,\ a>b>c>d$$
$a,b,c,d$ squared is equal to the square of $630$, and $a$ is larger than $b$, and so forth.
$a,b,c,d$ is also of such form that:
$$n = x + (x+1) + (x+2) + (x+3), \ x\in\mathbb{N}$$
where $n$ could be substituted for $ a, b, c \ $or $d$.
In what ways could one attack this problem?

Since $a, b, c, d$ are all of the form $4x+6$ for some $x \in \mathbb{N}$, we see that $a,b,c,d \equiv 2 \pmod 4$. It follows that $a^2+b^2+c^2+d^2 \equiv 4+4+4+4 \equiv 0 \pmod 8$. However, $630^2 \equiv 4 \pmod 8$. Therefore there is no solution.
