Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$ 
Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$.

Rearranging the given equation gives $k = \dfrac{n(n+1)}{m}-m$. Thus, $m \mid n(n+1)$ and so we have cases.
Case 1: $n = am$ for some $a \in \mathbb{Z}^+$
In this case we have $k = a(am+1)-m$, which is positive for all $a,m$.
Case 2: $n+1= am$ for some $a \in \mathbb{Z}^+$
In this case we have $k = a(am-1)-m$ and we need $k$ to be positive and so $a(am-1)-m \geq 1$, or $am \geq m+1$. This is true if and only if $a \geq 2$.
I didn't see how to solve the other case where both are not multiples of $m$.
 A: The equation
$$m(m+k) = n(n+1)$$
has positive integer solutions $m,n$ for all positive integers $k$ except $k = 2,3$.

Proof:

If $k = 1$, just let $m = n$, where $n$ is an arbitrary positive integer.

If $k$ is even, $k > 2$, let
$$m = (k(k-2))/4$$
$$n = ((k+2)(k-2))/4$$
and if $k$ is odd, $k > 3$, let
$$m = ((k-1)(k-3))/8$$
$$n = ((k+3)(k-3))/8$$
In both cases, the equation  $m(m+k) = n(n+1)$ is identically satisfied.

Next, suppose $k = 2$. Then
$$m(m + 2) = n(n + 1)$$
implies $m < n$ and $m > n-1$, contradiction.

Finally, suppose $k = 3$. Then
$$m(m + 3) = n(n + 1)$$
implies $m < n$ and $m > n -2$, hence $m = n -1$. But then
$$m(m + 3) = n(n+1) \implies (n-1)(n+2) = n(n+1) \implies n = \tfrac{1}{2}$$
contradiction. 

This completes the proof.
A: For each positive integer $N$, let $\tau(N)$ denote the number of positive integers $d$ that divides $N$.  For a given positive integer $u$, we also denote by $\bar{\tau}_u(N)$ the number of odd positive integers $d<u$ that divide $N$.  If $N=2^r\tilde{N}$, where $r\in\mathbb{Z}_{\geq 0}$ and $\tilde{N}\in\mathbb{Z}$ is odd, then write $\tilde{\tau}_u(N)$ the number of positive integers $d<u$ of the form $2^r\tilde{d}$, where $\tilde{d}$ is a divisor of $\tilde{N}$.  Obviously, $\bar{\tau}_1(N)=\tilde{\tau}_1(N)=0$ and $\bar{\tau}_u(N)\geq 1$ for $u>1$.
We claim that, for a given positive integer $k>1$, the number of solutions $(m,n)\in\mathbb{Z}_{>0}\times\mathbb{Z}_{>0}$ to the Diophantine equation
$$m(m+k)=n(n+1)$$
is
$$s_k:=\left\{
\begin{array}{ll}
\bar{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)+\tilde{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)&\text{if }k\text{ is odd}\,,
\\
\dfrac{1}{2}\,\tau(k^2-1)-1&\text{if }k\text{ is even}\,.
\end{array}\right.$$
Moreover, we set $s_1:=\infty$.  In particular, $s_k=0$ if and only if $k=2$ or $k=3$.  For an even integer $k>3$, there exists a unique solution if and only if $(k-1,k+1)$ is a pair of twin primes.  For an odd integer $k>3$, there exists a unique solution if and only if $k=7$ or $k=2F-1$, where $F$ is a Fermat prime (i.e., $F$ is a prime such that $F=2^{2^{l}}+1$ for some nonnegative integer $l$).
When $k=1$, there are $s_1=\infty$ solutions to the case $k=1$ given by $m=n$.  Suppose from now on that $k>1$.  Clearly, we hae $m<n$.  Furthermore,
$$(2n+1)^2< (2n+1)^2+(k^2-1)=4n(n+1)+k^2=4m(m+k)+k^2=(2m+k)^2\,.$$
That is, $2n+2\leq 2m+k$.  If $m=n-t$, then we have
$$1\leq t \leq \frac{k}{2}-1\,.$$
(At this stage, it is clear that, for a solution to exist, we must have $\dfrac{k}{2}-1\geq 1$, or $k\geq 4$.)
We then rewrite $m(m+k)=n(n+1)$ as
$$n(k-2t-1)=t(k-t)\,.$$
If $k$ is odd, we have
$$2n\,\left(\frac{k-1}{2}-t\right)=t(k-t)\,.$$
That is, since $t\equiv \frac{k-1}{2}\pmod{\frac{k-1}{2}-t}$, we have
$$\dfrac{k-1}{2}-t \,\mid\,\dfrac{k^2-1}{4}\,.$$
Indeed, for any divisor $d\in\mathbb{Z}_{>0}$ of $\dfrac{k^2-1}{4}$ such that $d<\dfrac{k-1}{2}$, we take $t:=\dfrac{k-1}{2}-d$, $$n:=\frac{\left(\frac{k-1}{2}-d\right)\left(\frac{k+1}{2}+d\right)}{2d}\text{ and }m:=n-t=n-\frac{k-1}{2}+d$$
to obtain a solution.  Now, note that $n$ defined in this way is a positive integer if and only if $d$ is odd, or $d$ is a maximally even divisor of $\dfrac{n^2-1}{4}$ (that is, $d$ is an even divisor such that $2d$ is not a divisor).  There are $s_k=\bar{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)+\tilde{\tau}_{\frac{k-1}{2}}\left(\dfrac{k^2-1}{4}\right)$ possible values of $d$.
If $k$ is even, we have
$$4n\,(k-2t-1)=(2t)\,(2k-2t)\,.$$
Note that $2t\equiv k-1\pmod{k-2t-1}$, so 
$$k-2t-1\,\mid\,k^2-1\,.$$
Indeed, let $d\in\mathbb{Z}_{>0}$ be a divisor of $k^2-1$ such that $d<k-1$.  We obtain a solution by setting $t:=\dfrac{k-1-d}{2}$,
$$n:=\frac{(k-1-d)(k+1+d)}{4d}\,,\text{ and }m:=n-t=n-\frac{k-1-d}{2}\,.$$
There are $s_k=\frac{1}{2}\,\tau(k^2-1)-1$ values of $d$.
The first odd value of $k>1$ with $s_k>1$ is $k=11$, where we have $s_{11}=3$ solutions $$(m,n)=(1,3)\,,\,\,(m,n)=(3,6)\,,\text{ and }(m,n)=(10,14)\,.$$  The first even value of $k>1$ with $s_k>1$ is $k=8$, where we have $s_8=2$ solutions $$(m,n)=(2,4)\text{ and }(m,n)=(12,15)\,.$$  In both odd and even cases, quasi's solutions correspond to $d=1$.
