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?
 A: In general
$$\frac{1}{u} + \frac{1}{v} = \frac{1}{n}, u,v,n \gt 0$$
can be solved as
$$n(u+v) = uv \iff (u-n)(v-n) = n^2$$
Thus you pick any factorization of $n^2$ and add $n$ to both factors.
If $n$ is odd, you get $u$ and $v$ to both be even. 
So to solve your equation, you can solve the one I mentioned here, and all you need is to find $n,u,v$ such that $u = v = 2 \mod 4$ and $n = 1 \mod 2$.
(For instance, you can pick $n$ to be a product of primes of the form $4k+1$)
A: $$\frac{1}{2a+1}+\frac{1}{(2a+1)(4a+1)}=\frac{2}{4a+1}$$
where $a>0$.
A: $\dfrac{1}{2m+1} +\dfrac{1}{2n+1}$, 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?
A: $$\frac{1}{2m+1}+\frac{1}{2n+1}=\frac{2n+1+2m+1}{(2m+1)(2n+1)}=\frac{2(n+m+1)}{(2m+1)(2n+1)}$$
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:
$$\frac{1}{2k+1}+\frac{1}{(2k+1)(4k+1)}=\frac{2}{4k+1}$$
