2013 Mock AIME I Problems/Problem 14

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

By Vieta's Formulas, the product of the roots is $-2014^2$ . Since $997$ is prime with $997\nmid2014^2$ , all the roots are relatively prime to $997$ . Thus, by Fermat's Little Theorem, we have $a_1^{997}+a_2^{997}+\cdots + a_{2013}^{997} \equiv a_1+a_2+\cdots + a_{2013} \pmod{997}$ , which, by Vieta, equals $-4 \equiv 993 \pmod{997}$ . Thus our answer is $\boxed{993}$ .

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2013 Mock AIME I Problems/Problem 14

Problem

Solution

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