HMMT February

Algebra Difficulty 5.1 AIME, harder Find the answer

It can be shown that there exists a unique polynomial $P$ in two variables such that for all positive integers $m$ and $n$ , $P(m, n)=\sum_{i=1}^{m} \sum_{j=1}^{n}(i+j)^{7}$ Compute $P(3,-3)$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Note that for integers $m>0, n>1$ , $P(m, n)-P(m, n-1)=\sum_{i=1}^{m}(i+n)^{7}=(n+1)^{7}+(n+2)^{7}+(n+3)^{7}$ for all real $n$ . Moreover, $P(3,1)-P(3,0)=P(3,1) \Longrightarrow P(3,0)=0$ . Then $\begin{aligned} P(3,-3) & =P(3,0)-\left(1^{7}+2^{7}+3^{7}\right)-\left(0^{7}+1^{7}+2^{7}\right)-\left((-1)^{7}+0^{7}+1^{7}\right) \\ & =-3^{7}-2 \cdot 2^{7}-2=-2445 \end{aligned}$

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