<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] $$\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2n+1+2m+1}{(2m+1)(2n+1)}=\frac{2(n+m+1)}{(2m+1)(2n+1)}$$
So this is of the form $\frac{2}{2k+1}$ if and only if $$(n+m+1)\mid(2m+1)(2n+1)$$
Now $(2n+1)(2m+1)=4nm + 2(n+m) + 1 = 4nm -1 + 2(n+m+1)$. So we want:
$$(n+m+1)\mid 4nm-1$$
Let $d=n+m+1$. Then $n\equiv -1-m\pmod d$ and therefore:
$$0\equiv 4nm-1\equiv 4(-1-m)m-1=-(2m+1)^2\pmod d$$
So $n+m+1\mid(2m+1)^2$, and, by symmetry, and $n+m+1\mid(2n+1)^2$.
Now assume $n<m$ and fix some $n\geq 1.$ We can find an $m$ if and only if there is a $d\mid (2n+1)^2$ such that $2n+1<d\leq(2n+1)^2$.  In that case $m=d-n-1$. 
If $(m,n)$ is a pair satisfying this condition and $m>n$ then there is some factoring $(2n+1)^2=PQ$ with $P>Q>1$ and $m=P-n-1$.
So we are basically talking about factorings of odd squares in non-trivial fashion.
The general solution for this is $P=a^2c$ and $Q=b^2c$ with $a>b$, with $a,b,c$ odd. This yields the general solution, with $a,b,c$ odd and $a>b$:
$$2n+1=abc,2m+1=a(2a-b)c,2k+1=bc(2a-b)$$
The base case is when $c=1$ and $a>b$ are odd and relatively prime. Then $n=\frac{ab-1}{2},m=\frac{a(2a-b)-1}{2},k=\frac{b(2a-b)-1}{2},$ and we get: $$\begin{align}\frac{1}{2n+1}+\frac{1}{2m+1}&=\frac{1}{ab}+\frac{1}{a(2a-b)}\\
&=\frac{2a-b+b}{a(2a-b)b}\\
&=\frac{2}{b(2a-b)}\\
&=\frac{2}{2k+1}\end{align}$$
All other solutions can be found by dividing a base solutions by an odd integer.
When $a=2k+1$ and $b=1,$ you get the selected answer:
$$\frac{1}{2k+1}+\frac{1}{(2k+1)(4k+1)}=\frac{2}{4k+1}$$
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