<s>[INST] 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. 
 [/INST] I'm not sure why you didn't simply finish it:
$8 = 2^2 + 2^2 +0^2 + 0^2$
So $(2^2 + 2^2 +0^2 + 0^2)(4^2 + 4^2 + 6^2 + 3^2)$
$= (2*4 - 2*4 - 0*6 - 0*3)^2 + (2*4+2*4+0*3-0*6)^2+(2*6 - 2*3 + 0*4 + 0*4)^2 + (2*3 + 2*6 - 0*4 + 0*4)^2$
$= 0^2 + 16^2 + 6^2 + 18^2$
$= 256+ 36 + 324 = 616$
</s>