Olympiad Mathematical Kosovo 2012 (Problem grade 9)

This is very easy to solve once you know congruences, and harder when you are young and do not. That is to say a "tricky", though nice, kind of problem.

There are 2012 - 29 = 1983 squares that are all 1 mod 3, since squares can only be 0 or 1 mod 3 (if a = 2 mod 3 then a^2 = 4 mod 3 = 1 mod 3) and we have excluded all those who were divisible by 3. Their sum will be equal to 1983 mod 3 = 0 mod 3, and so divisible by 3 since 1983 is divisible by 3.


Hint $\, $ mod prime $p:\ a\not\equiv 0\,\Rightarrow\, a^{p-1}\!\equiv 1,\,$ so $\,a_1^{p-1}\! +\cdots+ a_n^{p-1}\! \equiv\:$the number of $a_i\! \not\equiv 0$

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