Solve $(l+m)^{(l+m)}=(m(l+1))^m\cdot m^l$ While trying to answer a question from this site I arrived at this equation, which I should solve in natural numbers $m,l$ : $$(l+m)^{(l+m)}=(m(l+1))^m\cdot m^l$$.
I am not sure did I ever needed to solve something like this, and, honestly, and truthfully, I do not have a nice idea that should work.
Do you have any idea on how to solve this?
Should there be a finite number of solutions?
 A: Rewriting the equation for $m>0, l>0$ as
$$(l+1)^m=(1+\frac{l}{m})^{l+m}$$
we conclude that $l/m$ must be an integer. If it was an irreducible non-integer fraction then we can easily prove that its integer powers are also non-integer. 
we substitute back $l=nm, n\in \mathbb{N}$ and we obtain that
$$(n+1)^{n+1}=1+nm$$
Here, we can solve for m and we find that
$$m=\frac{(n+1)^{n+1}-1}{n}=\sum_{i=0}^{n}(n+1)^i$$
which is indeed a natural number, and there are no more constraints on it. Thus the full set of solutions can be found by varying $n$ on the integers:
$$m=\frac{(n+1)^{n+1}-1}{n}\\
l=(n+1)^{n+1}-1$$
To complete the classification, we notice that for $l=0$ any natural number $m\geq 0$ solves the equation, but for $m=0, l\neq 0$ there are no solutions.
A: $$(l+m)^{l+m}=(m(l+1))^m\cdot m^l=m^{l+m}(l+1)^m$$
Suppose $m=0$, then the equation simplifies to $l^l=0^l$, which has the sole solution $l=0$. Now assume $m>0$ and divide by $m^{l+m}$:
$$\left(1+\frac lm\right)^{l+m}=(l+1)^m$$
From this it is clear that $m\mid l$. Write $l=km$, $k$ being another integer:
$$(1+k)^{(k+1)m}=(km+1)^m$$
Both sides are now perfect $m$th powers, so take the $m$th root:
$$(1+k)^{k+1}=km+1$$
Since $(1+k)^{k+1}\equiv1^{k+1}\equiv1\equiv km+1\bmod k$, any $k>0$ yields a unique solution $s(k)=\frac{(1+k)^{k+1}-1}k$ for $m$. When $k=0$, $m$ may be any number whatsoever, but then $l=0$. Therefore, the solutions $(l,m)$ to the original equation are described by $\{(k\cdot s(k),k):k>0\}\cup\{(0,m):m\ge0\}$.
A: I replace $l$ by $n$,
so the equation becomes
$(n+m)^{n+m}
=(m(n+1))^m m^n
$.
I will derive bounds
for $m$ in terms of $n$,
and leave it at that
since I have more pressing
things to do.
$(n+m)^{n+m}
=(m(n+1))^m m^n
=(n+1)^m m^{n+m}
$
or
$(1+n/m)^{n+m}
=(n+1)^m
$
or
$(1+n/m)^{1+n/m}
=(n+1)
$.
Therefore
$x^x = n+1$
where
$x = 1+n/m$.
Write as
$x\ln(x) = \ln(n+1)
$.
Let
$x = e^y, \ln(n+1) = t$,
so that
$ye^y = t$.
By De Bruijn,
"Asymptotic Methods in Analysis",
 2.4.4,
$y
=L_1-L_2+L_2/L_1+\frac12 (L_2-1)L_2/L_1^2 + ...
$
where
$L_1 = \ln(t),
L_2 = \ln\ln(t)
$.
In particular,
$L_1-L_2
\lt y
\lt L_1
$
so
$t/\ln(t)
\lt x
\lt t
$
or
$\ln(n+1)/\ln(\ln(n+1))
\lt 1+n/m
\lt \ln(n+1)
$.
From the right,
$m
\gt \dfrac{n}{\ln(n+1)-1}
$
and,
from the left,
$m
\lt \dfrac{n}{\ln(n+1)/\ln(\ln(n+1))-1}
$.
