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NT about the positive divisors of n and n+1

Source: Turkey EGMO TST 2019 #4

July 15, 2020
number theorynumber theory proposed

Problem Statement

Let σ(n)\sigma (n) shows the number of positive divisors of nn. Let s(n)s(n) be the number of positive divisors of n+1n+1 such that for every divisor aa, a−1a-1 is also a divisor of nn. Find the maximum value of 2s(n)−σ(n)2s(n)- \sigma (n) .
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