2013 Mock AIME I Problems/Problem 14

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Problem

Let $P(x) = x^{2013}+4x^{2012}+9x^{2011}+16x^{2010}+\cdots + 4052169x + 4056196 = \sum_{j=1}^{2014}j^2x^{2014-j}.$ If $a_1, a_2, \cdots a_{2013}$ are its roots, then compute the remainder when $a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997}$ is divided by 997.

Solution

Since 997 is prime, we have $a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997}$ equals to $a_1+a_2+\cdots + a_{2013}$ mod 997, which by Vieta's equals -4. Thus our answer is 993 mod 997.

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