Writing 616 as a sum of four squares I am using the fact that $S_4$ is closed under multiplication and Euler's four square identity:
$$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2) =\\ 
\quad(a_1b_1 - a_2b_2 - a_3b_3 - a_4b_4)^2 + (a_1b_2+a_2b_1+a_3b_4-a_4b_3)^2
+(a_1b_3 - a_2b_4 + a_3b_1 + a_4b_2)^2 + (a_1b_4 + a_2b_3 - a_3b_2 + a_4b_1)^2$$
So $616= 2^3\cdot 7\cdot 11$
$= 8 \times (2^2+1^2+1^2+1^2)(3^2+1^2+1^2+0^2)$
$=8 \times(6 − 1 − 1)^2 + (2 + 3 + 0 − 1)^2 + (2 − 0 + 3 + 1)^2 + (0 + 1 − 1 + 3)^2$
$=8(4^2+4^2+6^2+3^2)$
But then from here I am stuck, I can't see how to get the four integers. 

If the sum of four squares is divisible by $8,$ then each of the four numbers is even and we may divide through by $2.$ That is, we write $616/4 = 154$ as the sum of four squares (this allows $0$ and $1$ as squares if convenient). Then just double those. 
$$ 154 = 144 + 9 + 1 + 0 $$
$$ 154 = 121 + 25 + 4 + 4 $$
$$ 154 = 121 + 16 + 16 + 1   $$
and so on
