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Algebra Difficulty 5.3 AIME, harder Find the answer

14. Determine the number of ordered pairs of integers (m,n)(m, n) for which mn≥0m n \geq 0 and m3+n3+99mn=333m^{3}+n^{3}+99 m n=33^{3}.
(2 marks)
求滿足 mn≥0m n \geq 0 及 m3+n3+99mn=333m^{3}+n^{3}+99 m n=33^{3} 的整數序偶 (m,n)(m, n) 的數目。
(2 分)

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

Solution

14. 35
14. Using the identity
a3+b3+c3−3abc=12(a+b+c)[(a−b)2+(b−c)2+(c−a)2], a^{3}+b^{3}+c^{3}-3 a b c=\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right],
we have
m3+n3+99mn−333=m3+n3+(−33)3−3mn(−33)=12(m+n−33)[(m−n)2+(m+33)2+(n+33)2] \begin{aligned} m^{3}+n^{3}+99 m n-33^{3} & =m^{3}+n^{3}+(-33)^{3}-3 m n(-33) \\ & =\frac{1}{2}(m+n-33)\left[(m-n)^{2}+(m+33)^{2}+(n+33)^{2}\right] \end{aligned}

For this expression to be equal to 0 , we either have m+n=33m+n=33 or m=n=−33m=n=-33. The latter gives one solution (−33,−33)(-33,-33) while the former gives the 34 solutions (0,33),(1,32),…,(33,0)(0,33),(1,32), \ldots,(33,0). Hence the answer is 35 .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.