The rest of $a^2+b^2$

Lazy solution

Obviously, $(a,b) = (0,0)$ is a solution. If this problem has a unique answer, as suggested, then it is $0^2 + 0^2 = 0 \equiv 0\ \text{mod}\ 2002$ .

Oh, you say, I need to prove it's the only correct answer? Ah.

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Complete solution

Let $a + 5b = M$ and $5a - b = N$ . Then $a = \frac{M + 5N}{26},\ \ \ \ b = \frac{5M - N}{26}.$ When substituting this into $a^2 + b^2$ , then cross-terms with $MN$ cancel, and we are left with $a^2 + b^2 = \frac{26M^2 + 26N^2}{26^2} = \frac{M^2 + N^2}{26}.$ Since $M$ and $N$ are multiples of 2002, the numerator is a multiple of $2002^2 = 2002\cdot 77\cdot 26$ ; after division by 26, the result is divisible by $2002\cdot 77$ , so that division by 2002 will give remainder $\boxed{0}$ .