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

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 $993\pmod{997}$.

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS