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Number theory Difficulty 5.1 AIME, harder Prove it Soviet Union

Problem:
Given unequal integers xx, yy, zz prove that (xy)5+(yz)5+(zx)5\left(x - y\right)^{5} + \left(y - z\right)^{5} + \left(z - x\right)^{5} is divisible by 5(xy)(yz)(zx)5(x - y)(y - z)(z - x).

Solution

Solution:
Put xy=rx - y = r, yz=sy - z = s. Then zx=(r+s)z - x = - (r + s), and
(xy)5+(yz)5+(zx)5=r5+s5(r+s)5 \left(x - y\right)^{5} + \left(y - z\right)^{5} + \left(z - x\right)^{5} = r^{5} + s^{5} - (r + s)^{5}
Expand (r+s)5(r + s)^5:
(r+s)5=r5+5r4s+10r3s2+10r2s3+5rs4+s5 (r + s)^5 = r^5 + 5r^4s + 10r^3s^2 + 10r^2s^3 + 5rs^4 + s^5
So,
r5+s5(r+s)5=r5+s5[r5+5r4s+10r3s2+10r2s3+5rs4+s5] r^5 + s^5 - (r + s)^5 = r^5 + s^5 - [r^5 + 5r^4s + 10r^3s^2 + 10r^2s^3 + 5rs^4 + s^5]
=5r4s10r3s210r2s35rs4 = -5r^4s - 10r^3s^2 - 10r^2s^3 - 5rs^4
=5rs(r3+2r2s+2rs2+s3) = -5rs(r^3 + 2r^2s + 2rs^2 + s^3)
=5rs(r2+rs+s2)(r+s) = -5rs(r^2 + rs + s^2)(r + s)
Therefore,
(xy)5+(yz)5+(zx)5=5rs(r+s)(r2+rs+s2) \left(x - y\right)^{5} + \left(y - z\right)^{5} + \left(z - x\right)^{5} = -5rs(r + s)(r^2 + rs + s^2)
Recall r=xyr = x - y, s=yzs = y - z, r+s=xy+yz=xz=(zx)r + s = x - y + y - z = x - z = -(z - x).

Thus, rs(r+s)=(xy)(yz)(zx)rs(r + s) = (x - y)(y - z)(z - x) (up to sign), and the expression is divisible by 5(xy)(yz)(zx)5(x - y)(y - z)(z - x).

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.