How can we show that for any integer $k>1$ there are positive integers $m$ and $n$ such that $\frac{1}{k}=\frac{m-n}{(m+1)(n+1)}.$
(Thanks to Arthur Fischer for the reformulation!)
Proving $\frac{m-n}{(m+1)(n+1)}=\frac{1}{k}$ for every k>1
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elementary-number-theory
diophantine-equations
1 Answers
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The claim is false for $k=2$. For $k$ not $2$, $\frac{1}{k}=\frac{(k(k-1)-1)-(k-2)}{k(k-1)(k-1)}$.