olymmath

TSTST 2020 Problem 8

    For every positive integer NN, let σ(N)\sigma(N) denote the sum of the positive integer divisors of NN. Find all integers m≥n≥2m \geq n \geq 2 satisfying σ(m)−1m−1=σ(n)−1n−1=σ(mn)−1mn−1.\frac{\sigma(m)-1}{m-1}=\frac{\sigma(n)-1}{n-1}=\frac{\sigma(m n)-1}{m n-1} .

    Dataset text NuminaMath olympiads. Quoted with source.

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