Show that from a set of all 11 integers one can select six numbers such that $a^2+b^2+c^2=d^2+e^2+f^2\pmod {12} $) I have got an INMO problem(2015), which is as follow:

Show that from any set of $11$ integers one can select six numbers such that $a^2+b^2+c^2\equiv d^2+e^2+f^2\pmod{12} $.

Here we have freedom to choose $11$ arbitrary integers and $a, b, c, d, e, f$ are 6 of those $11$ integers.
I m clueless. I don't know even what should be my approach to start the question. Any hint or solution is heartily welcome.Thanks.
 A: The squares modulo $12$ are $0,1,4,9$.
Partition your set $V$ of $11$ integers into $V_0,V_1,V_4,V_9$ with $$V_k=\{x\in V\mid x^2\equiv k\pmod{12}\}$$
Sort $0,1,4,9$ to $k_1,k_2,k_3,k_4$ so that $|V_{k_1}|\geq |V_{k_2}|\geq |V_{k_3}|\geq |V_{k_4}|$
If three of the $V_i$ have two or more elements, then you have a solution.
So this means $|V_{k_3}|\leq 1$ and $|V_{k_3}\cup V_{k_4}|\leq 2$.
This means $|V_{k_1}|\geq 5$. If $|V_{k_1}|\geq 6$ then you have a solution of the form $k_1+k_1+k_1=k_1+k_1+k_1$.
So $|V_{k_1}|=5$ and $|V_{k_2}|=4,5$. But that means we have a solution of the form: 
$$k_1+k_1+k_2=k_1+k_1+k_2$$
This seem to be true for $9$ integers, as well. If $|V_{k_1}|\geq 6$ or $|V_{k_1}|\geq 4$ and $|V_{k_2}|\geq 2$ or $|V_{k_1}|,|V_{k_2}|,|V_{k_3}|\geq 2$ there are solutions.
For eight integers, you can choose $|V_1|=5,|V_0|=1,|V_4|=1,|V_{9}|=1$

We have actually proved more specifically that there are six elements $a,b,c,d,e,f$ such that:
$$a^2\equiv d^2\pmod{12}\\
b^2\equiv e^2\pmod{12}\\
c^2\equiv f^2\pmod{12}$$
This is easier to see directly by noting that amongst $5$ integers, we can always choose a pair $x,y$ so that $x^2\equiv y^2\pmod{12}$.
Then given nine integers, we pick $a,d$, leaving $7$ integers, from which we select $b,e$, leaving $5$ integers, from which we select $c,f$.
