Maths Olympiad Prep

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Number theory Difficulty 5.8 AIME, harder Prove it

14. Below are various forms of the Möbius inversion formula, which can be directly verified or derived using dnμ(d)=[1n]\sum_{d \mid n} \mu(d)=\left[\frac{1}{n}\right]:
(i) Let x1,Kx \geqslant 1, K be a given positive integer, and let 1nx,nK1 \leqslant n \leqslant x, n \mid K. Prove that F(n)=dKndxf(d)F(n) = \sum_{\substack{d|K \\ n| d \leqslant x}} f(d) holds if and only if f(n)=dKndxμ(dn)F(d)f(n) = \sum_{\substack{d|K \\ n| d \leqslant x}} \mu\left(\frac{d}{n}\right) F(d).
(ii) Let α(x),β(x)\alpha(x), \beta(x) be functions defined on the interval (0,)(0, \infty). Prove that β(x)=d=1α(xd)\beta(x) = \sum_{d=1}^{\infty} \alpha\left(\frac{x}{d}\right) holds if and only if α(x)=d=1μ(d)β(xd)\alpha(x) = \sum_{d=1}^{\infty} \mu(d) \beta\left(\frac{x}{d}\right), assuming that for a given x>0x > 0, the double series d=1k=1α(xdk)\sum_{d=1}^{\infty} \sum_{k=1}^{\infty}\left|\alpha\left(\frac{x}{d k}\right)\right| and d=1k=1β(xdk)\sum_{d=1}^{\infty} \sum_{k=1}^{\infty}\left|\beta\left(\frac{x}{d k}\right)\right| both converge.
(vi) Let α(x,y),β(x,y)\alpha(x, y), \beta(x, y) be functions of two real variables x,yx, y defined on the rectangular region 0<x0xx1,0<y0yy10 < x_{0} \leqslant x \leqslant x_{1}, 0 < y_{0} \leqslant y \leqslant y_{1}. Prove that
β(x,y)=1dx1/x1ly1/yα(dx,ly)\beta(x, y) = \sum_{1 \leqslant d \leqslant x_{1} / x} \sum_{1 \leqslant l \leqslant y_{1} / y} \alpha(d x, l y)
holds if and only if
α(x,y)=1dx1/x1ly1/yμ(d)μ(l)β(dx,ly).\alpha(x, y) = \sum_{1 \leqslant d \leqslant x_{1} / x} \sum_{1 \leqslant l \leqslant y_{1} / y} \mu(d) \mu(l) \beta(d x, l y).
(vii) Analogous to the generalization of (ii) to (iii), (iv), (v), make the corresponding generalization for (vi).

Solution

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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.