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Algebra Difficulty 5.1 AIME, harder Find the answer

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

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that for integers m>0,n>1m>0, n>1, P(m,n)−P(m,n−1)=∑i=1m(i+n)7=(n+1)7+(n+2)7+(n+3)7P(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 nn. Moreover, P(3,1)−P(3,0)=P(3,1)⟹P(3,0)=0P(3,1)-P(3,0)=P(3,1) \Longrightarrow P(3,0)=0. Then P(3,−3)=P(3,0)−(17+27+37)−(07+17+27)−((−1)7+07+17)=−37−2⋅27−2=−2445\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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