show that 2016 can be written as a sum of 4 squares

Question

Show that 2016 can be written as a sum of 4 squares. My Try:- As $2016=40^2+4^2+16^2+12^2$, we can prove the above result. But I need actual/real process.

Answer 1

Odd squares are all congruent to $1$ mod $8$, while even squares are congruent to $0$ or $4$ mod $8$. Consequently, the sum of four odd squares is congruent to $4$ mod $8$, while the sum of two odd and two even squares is congruent to $2$ or $6$ mod $8$. Since $2016\equiv0$ mod $8$, it can therefore only be the sum of four even squares, i.e., $2016=(2x)^2+(2y)^2+(2z)^2+(2w)^2$. (It cannot be sum the of one odd and three even, or three even and one odd, squares, for mod $2$ reasons.) This reduces to $504=x^2+y^2+z^2+w^2$. But since $504\equiv0$ mod $8$ also, the same reasoning applies again, so that we must have $504=(2a)^2+(2b)^2+(2c)^2+(2d)^2$, which reduces to $126=a^2+b^2+c^2+d^2$. This is now small enough that all solutions are easily found, i.e.,

$$126= \begin{cases}121+4+1+0\\ 100+25+1+0\\ 100+16+9+1\\ 81+36+9+0\\ 81+25+16+4\\ 64+49+9+4\\ 64+36+25+1\\ 49+36+25+16 \end{cases}$$

Taking $a$, $b$, $c$ and $d$ (in any order) from any of these solutions gives $2016=(4a)^2+(4b)^2+(4c)^2+(4d)^2$, e.g, $44^2+8^2+4^2+0^2$ (from the top) or $28^2+24^2+20^2+16^2$ (from the bottom).

Answer 2

First notice that if $n=a^2+b^2+c^2+d^2$, then $x^2n = (ax)^2+(bx)^2+(cx)^2+(dx)^2.$ So the first thing I'd do in this problem is factor 2016 and see what perfect squares I find. $2016 = 144\cdot 14$. Now express 14 as the sum of 4 squares: $14 = 3^2+2^2 + 1^2 +0^2$. Then $2016 = 36^2+24^2+12^2+0^0.$

If you don't like the $0^2$, then move some factor from the $144$ to the $14$: $2016 = 16\cdot 126$ and $126 = 10^2+4^2+3^2+1^2$. Multiply through by $4^2$ and this gives your answer.

We're lucky in this problem that there are lots of squares in $2016$ and that it's not too big.