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:

Let us take such 5 pairs: say (x²₁ ; y²₁); (x²₂ ; y²₂); .... (x²₅ ; y²₅). Then x²_j - y²_j is divisible by 4 for 1 ≤ j ≤ 5. 

Let r_j be the remainder when x²_j - y²_j is divisible by 3, 1 ≤ j ≤ 3. 

We have 5 remainders r₁; r₂; r₃; r₄; r₅. 

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 r₁ = r₂ = r₃, then 3 divides r₁ + r₂ + r₃; if r₁ = 0; r₂ = 1 and r₃ = 2, then again 3 divides r₁+r₂+r₃. 

Thus we can always find three remainders whose sum is divisible by 3. This means we can find 3 pairs, say, (x²₁ ; y²₁); (x²₂ ; y²₂); (x²₃ ; y²₃) such that 3 divides (x²₁ - y²₁) + (x²₂ - y²₂)+(x²₃ - y²₃). 

Since each difference is divisible by 4, we conclude that we can find 6 numbers a²; b²; c²; d²; e²; f² such that 

a² + b² + c² ≡ d² + e² + f² (mod 12):