<s>[INST] Find this $\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2}{2k+1}$ Let  $m,n,k\in \mathbb{N}$, and $m,n,k\ge 1,m\neq n\neq k\neq m $, such that
$$
\dfrac{1}{2m+1}+\dfrac{1}{2n+1}=\dfrac{2}{2k+1}
$$
Is there a solution? Or does this not have any solution?
 [/INST] $\dfrac{1}{2m+1} +\dfrac{1}{2n+1}$, letting $2m+1$ to be the LCM of $2m+1$ and $2n+1$, Assume $2m+1=(2n+1) \cdot k$, your expression becomes $\dfrac{k+1}{2m+1}=\dfrac{2}{2l+1}$
$\dfrac{(k+1)}{2m+1}=\dfrac{2}{2l+1} \implies k+1=\dfrac{2(2m+1)}{2l+1}$. You just need a condition that $2l+1|2m+1$. Do you see that infinitely many solutions exists for this?
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