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[数论] 某分式不定方程正整数解数

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青青子衿 Posted 2021-4-27 11:47 |Read mode
给定正整数$n$,分式不定方程$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{xy}=\dfrac{1}{n}$的正整数解数等于$n(n+1)$因子个数.
  1. Table[{Length[
  2.    Solve[1/x + 1/y + 1/(x*y) == 1/k && 0 < {x, y}, {x, y}, Integers]],
  3.    DivisorSigma[0, k (k + 1)]}, {k, 1, 100}]
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kuing Posted 2021-4-27 13:38
?去分母整理一下不就变成 `(x-n)(y-n)=n(n+1)` 么……

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