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Number theory Difficulty 5.5 AIME, harder Prove it

4・260 Proof: The equation
x2+y2+z2=x3+y3+z3x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}

has infinitely many integer solutions.

Solution

［Proof］Let
{x=n(4n2−1)+1,y=1−n(4n2−1),z=1−4n2n=1,2,⋯ \left\{\begin{array}{l} x=n\left(4 n^{2}-1\right)+1, \\ y=1-n\left(4 n^{2}-1\right), \\ z=1-4 n^{2} \end{array} \quad n=1,2, \cdots\right.

Then, we have
x2(x−1)=[n(4n2−1)+1]2⋅n(4n2−1)=[n(4n2−1)−1]2⋅n(4n2−1)+(4n2−1)2.4n2=y2(1−y)+z2(1−z)\begin{aligned} x^{2}(x-1)= & {\left[n\left(4 n^{2}-1\right)+1\right]^{2} \cdot n\left(4 n^{2}-1\right) } \\ = & {\left[n\left(4 n^{2}-1\right)-1\right]^{2} \cdot n\left(4 n^{2}-1\right)+\left(4 n^{2}-1\right)^{2} . } \\ & 4 n^{2} \\ = & y^{2}(1-y)+z^{2}(1-z) \end{aligned}

That is, x3+y3+z3=x2+y2+z2x^{3}+y^{3}+z^{3}=x^{2}+y^{2}+z^{2}.
Therefore, the original equation has infinitely many integer solutions.

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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.