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Algebra Difficulty 4.8 AIME Find the answer

Find the number of ordered triples of integers (a,b,c)(a, b, c) with 1≤a,b,c≤1001 \leq a, b, c \leq 100 and a2b+b2c+c2a=ab2+bc2+ca2a^{2} b+b^{2} c+c^{2} a=a b^{2}+b c^{2}+c a^{2}

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

Solution

This factors as (a−b)(b−c)(c−a)=0(a-b)(b-c)(c-a)=0. By the inclusion-exclusion principle, we get 3⋅1002−3⋅100+100=298003 \cdot 100^{2}-3 \cdot 100+100=29800.

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