3
$\begingroup$

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.

$\endgroup$
6
  • 7
    $\begingroup$ $8=2^2+2^2+0^2+0^2$. Couldn't you use the same identity again? $\endgroup$
    – Arthur
    Jun 2 '17 at 11:54
  • $\begingroup$ $16^2 + 14^2+8^2+10^2 = 616$ $\endgroup$ Jun 2 '17 at 12:17
  • 1
    $\begingroup$ shortcut is: $616=28\cdot22=(5^2+1^2+1^2+1^2)(4^2+2^2+1^2+1^2)$ $\endgroup$
    – farruhota
    Jun 2 '17 at 12:35
  • 1
    $\begingroup$ My strategy was to divide $616$ by $4$ and express $154$ as sum of three squares, achieving the same result as @GerryMyerson. $\endgroup$
    – Lubin
    Jun 2 '17 at 17:45
  • 1
    $\begingroup$ For reference, all solutions satisfying $0 \le a \le b \le c \le d$. $$\begin{array}{rll}616 &= 0^2 + 2^2 + 6^2 + 24^2 &= 0^2 + 6^2 + 16^2 + 18^2\\ &= 2^2 + 4^2 + 14^2 + 20^2 &= 2^2 + 8^2 + 8^2 + 22^2\\ &= 2^2 + 10^2 + 16^2 + 16^2 &= 2^2 + 12^2 + 12^2 + 18^2\\ &= 4^2 + 4^2 + 10^2 + 22^2 &= 4^2 + 10^2 + 10^2 + 20^2\\ &= 6^2 + 6^2 + 12^2 + 20^2 &= 8^2 + 10^2 + 14^2 + 16^2 \end{array}$$ this is generated by brute force, no idea how to exhaust all solutions in other manner. $\endgroup$ Jun 2 '17 at 19:32
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

I'm not sure why you didn't simply finish it:

$8 = 2^2 + 2^2 +0^2 + 0^2$

So $(2^2 + 2^2 +0^2 + 0^2)(4^2 + 4^2 + 6^2 + 3^2)$

$= (2*4 - 2*4 - 0*6 - 0*3)^2 + (2*4+2*4+0*3-0*6)^2+(2*6 - 2*3 + 0*4 + 0*4)^2 + (2*3 + 2*6 - 0*4 + 0*4)^2$

$= 0^2 + 16^2 + 6^2 + 18^2$

$= 256+ 36 + 324 = 616$

$\endgroup$
1
  • $\begingroup$ Of course that's not the only way to do it but is does use Eulers identity only. You could do any $(8*7)*11 = (8*11)*7 = 8*(7*11)$ as well. $\endgroup$
    – fleablood
    Jun 2 '17 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.