Problem 1038

Solution: The first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:

| Odd numbers | Even numbers | Odd pairs | Even pairs | Total pairs |
| :---: | :---: | :---: | :---: | :---: |
| 0 | 11 | 0 | 5 | 5 |
| 1 | 10 | 0 | 5 | 5 |
| 2 | 9 | 1 | 4 | 5 |
| 3 | 8 | 1 | 4 | 5 |
| 4 | 7 | 2 | 3 | 5 |
| 5 | 6 | 2 | 3 | 5 |
| 6 | 5 | 3 | 2 | 5 |
| 7 | 4 | 3 | 2 | 5 |
| 8 | 3 | 4 | 1 | 5 |
| 9 | 2 | 4 | 1 | 5 |
| 10 | 1 | 5 | 0 | 5 |
| 11 | 0 | 5 | 0 | 5 |

Let us take such 5 pairs: say $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right), \ldots,\left(x_{5}^{2}, y_{5}^{2}\right)$. Then $x_{j}^{2}-y_{j}^{2}$ is divisible by 4 for $1 \leq j \leq 5$. Let $r_{j}$ be the remainder when $x_{j}^{2}-y_{j}^{2}$ is divisible by $3,1 \leq j \leq 3$. We have 5 remainders $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. But these can be 0,1 or 2 . Hence either one of the remainders occur 3 times or each of the remainders occur once. If, for example $r_{1}=r_{2}=r_{3}$, then 3 divides $r_{1}+r_{2}+r_{3}$; if $r_{1}=0, r_{2}=1$ and $r_{3}=2$, then again 3 divides $r_{1}+r_{2}+r_{3}$. Thus we can always find three remainders whose sum is divisible by 3 . This means we can find 3 pairs, say, $\left(x_{1}^{2}, y_{1}^{2}\right),\left(x_{2}^{2}, y_{2}^{2}\right),\left(x_{3}^{2}, y_{3}^{2}\right)$ such that 3 divides $\left(x_{1}^{2}-y_{1}^{2}\right)+\left(x_{2}^{2}-y_{2}^{2}\right)+\left(x_{3}^{2}-y_{3}^{2}\right)$. Since each difference is divisible by 4 , we conclude that we can find 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that

$$a^{2}+b^{2}+c^{2} \equiv d^{2}+e^{2}+f^{2} \quad(\bmod 12)$$