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\tau (n^m) = \sum_{d|n} m^{\nu (d)}

Source: Mathcenter Contest / Oly - Thai Forum 2012 sl-7 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

November 13, 2022
number theory

Problem Statement

The arithmetic function ν\nu is defined by ν(n)={0,     n=1k,     n=p1a1p2a2...pkak\nu (n) = \begin{cases}0, \,\,\,\,\, n=1 \\ k, \,\,\,\,\, n=p_1^{a_1} p_2^{a_2} ... p_k^{a_k}\end{cases}, where n=p1a1p2a2...pkakn=p_1^{a_1} p_2^{a_2} ... p_k^{a_k} represents the prime factorization of the number. Prove that for any naturals m,nm,n, τ(nm)=∑d∣nmν(d).\tau (n^m) = \sum_{d | n} m^{\nu (d)}. (PP-nine)
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