Question

From a set of 11 square integers, show that one can choose 6 numbers $a^2,b^2,c^2,d^2,e^2,f^2$ such that $\left\{\right(a^2+b^2+c^2)-(d^2+e^2+f^2\left)\right\}$ is divisible by $12$.

Answer 1

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	1	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 $(x_1^2,y_1^2),(x_2^2,y_2^2),.......,(x_5^2,y_5^2)$.
Then $x_1^2-y_1^2$ is is divisible by 4 for $1\le j\le 5$.
Let $r_j$ be the remainder when $x_1^2-y_1^2$ is divisible by 3, $1\le j\le 3$.
There are 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 $r_1=0,r_2=1and{r}_3=2$, 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, $(x_1^2,y_1^2),(x_2^2,y_2^2),(x_3^2,y_3^2)$ such that 3 divides $(x_1^2,y_1^2)+(x_2^2,y_2^2)+(x_3^2,y_3^2)$.

Since each difference is divisible by 4, then $a^2,b^2,c^2,d^2,e^2,f^2$ such that $a^2+b^2+c^2{d}^2+e^2+f^2$ (mod 12).