Does the equation $q^2\, n(n+1) = p^2\, m(m+1)$ have solutions only if $m(m+1) = q^2\, k(k+1)$, $kI have the following equation
\begin{equation}
q^2\, n(n+1) = p^2\, m(m+1), \qquad(1)
\end{equation}
where $m,n,p,q \in \mathbb{N}$ and $p>q>2$ and $\text{gcd}(p,q)=1$.
I could find numerically solutions only when $m(m+1) = q^2\, k(k+1)$ and $n(n+1) = p^2\, k(k+1)$, where $k<m$, $k \in \mathbb{N}$. For example:


*

*$p=35$, $q=6$, $m=8$, $n=49$, so
$$m(m+1)=8\cdot 9=6^2 \cdot (1 \cdot 2)$$
$$n(n+1)=49 \cdot 50=35^2 \cdot (1 \cdot 2)$$

*$p=99$, $q=10$, $m=24$, $n=242$, so
$$m(m+1)=24\cdot 25=10^2 \cdot (2 \cdot 3)$$
$$n(n+1)=242\cdot 243 = 99^2 \cdot (2 \cdot 3)$$
Is it possible to prove (or disprove) that the equation (1) has solution only if $m(m+1) = q^2\, k(k+1)$ and $n(n+1) = p^2\, k(k+1)$, $k<m$ ?


Edit:
From eq. (1) and the given conditions it follows that $q^2 | m(m+1)$ and $p^2 | n(n+1)$. Therefore
\begin{align}
m(m+1) &= q^2 d, \qquad (2a)\\
n(n+1) &= p^2 d, \qquad (2b)\\
\end{align}
where $d = \text{gcd}(m(m+1),\, n(n+1))$. Since $m(m+1)$ and $n(n+1)$ are even numbers, it follows from (2) that $d$ is even non-square number. Numerically I could find solutions to the eq. (1) only if $d = k(k+1)$, where $k<m$, $k \in \mathbb{N}$.

First of all, a number $N$ is of the form $m(m+1)$ if and only if $4N+1$ is a perfect square.
Since $p$ and $q$ are coprime, we have $q^2|m(m+1)$, so $$\frac{m(m+1)}{q^2}$$ is a positive integer and we have $$S:=\frac{m(m+1)}{q^2}=\frac{n(n+1)}{p^2}$$
Using the criterion above, we get that $$4Sq^2+1$$ and $$4Sp^2+1$$ must be a perfect square, so the $y$-values of the pairs $(x/y)$ solving the Pell-equation $$x^2-4Sy^2=1$$ include $y=p$ and $y=q$. If the fundamental solution is $(a/b)$ it is easy to see that every $y$ must be a multiple of $b$. But $p$ and $q$ are coprime, hence $b=1$. This means that $4S+1$ is a perfect square, so $S=k(k+1)$ and of course $k<m$, completing the proof.
