The classic solution to show existence of one solution is to use the identity
$$ (n+3)^2 - (n+2)^2 - (n+1)^2 + n^2 = 4$$
and the fact that $1,2,3,4$ have a representation:
$$1 = + 1^2$$ $$2 = - 1^2 - 2^2 - 3^2 + 4^2$$ $$3 = -1^2 + 2^2$$ $$4 = -1^2 - 2^2 + 3^2$$
To get one representation of $4m+r$, we inductively get one representation for $4(m-1) + r$, and use the above identity.
Now, as Andre pointed out, given one representation we can extend that to infinitely many representations by writing $0+0+0 \dots$ as $(4 - 4) + (4-4) + (4-4) \dots$ and using the above identity multiple times.