Prove that if $d\cdot e| d(d+1)+e\cdot e$ then $d\cdot (d+1)+e\cdot e=3de$ Prove that if $d\cdot e| d(d+1)+e\cdot e$ then $d\cdot (d+1)+e\cdot e=3de$ where $d$ and $e$ are positive integers.

We can simplify the problem a bit. Let $\gamma = \gcd(d,e)$ and write $d = \gamma\delta,\; e = \gamma\varepsilon$. Then the condition becomes
$$\gamma^2\delta\varepsilon \mid \gamma\delta(\gamma\delta+1) + \gamma^2\varepsilon^2\tag{1}$$
and we see that we must have $\gamma \mid \delta$. So write $d = \gamma^2\kappa$ to obtain
$$\begin{align}
\gamma^3\kappa\varepsilon &\mid \gamma^2\kappa(\gamma^2\kappa+1) + \gamma^2\varepsilon^2\\
\iff \gamma\kappa\varepsilon &\mid \kappa(\gamma^2\kappa + 1) + \varepsilon^2.
\end{align}$$
Hence $\kappa \mid \varepsilon^2$. But $\kappa \mid \delta$ and $\gcd(\delta,\varepsilon) = 1$, therefore $\kappa = 1$ and we obtain
$$\gamma\varepsilon \mid \gamma^2 + 1 + \varepsilon^2.\tag{2}$$
That is equivalent to
$$\gamma \mid \varepsilon^2+1 \land \varepsilon \mid \gamma^2+1.\tag{3}$$
Since
$$F_{2k+1}^2 + 1 = F_{2k+3}F_{2k-1},$$
we have the family of consecutive odd-indexed Fibonacci numbers as solutions for $(\gamma,\,\varepsilon)$. For the original $d,\,e$, that means
$$d = F_{2k+1}^2,\quad \text{ and } e \in \{ F_{2k+1}F_{2k-1},\, F_{2k+1}F_{2k+3}\}.$$
It is easily verified that
$$\frac{\gamma^2+1+\varepsilon^2}{\gamma\varepsilon} = 3 = \frac{d(d+1)+e^2}{de}$$
for these pairs.
The above are the only solutions of $(3)$:
Let $1 < a < b$ with $a \mid b^2 + 1$ and $b \mid a^2 +1$. Write $a^2+1 = b\cdot c$ and $b^2+1 = a\cdot k$. Then $ 0 < c < a$ (since $a < b$), we have $c \mid a^2 + 1$, and
$$\begin{align}
c^2 + 1 &= \left(\frac{a^2+1}{b}\right)^2+1\\
&= \frac{(a^2+1)^2 + b^2}{b^2} = \frac{a^4 + 2a^2 + (1 + b^2)}{b^2}\\
&= \frac{a^2(a^2+2) + ak}{b^2}.
\end{align}$$
Since $\gcd(a,b) = 1$, the common factor $a$ in the numerator cannot be cancelled (even partially) by the $b^2$ in the denominator, hence also $a \mid c^2+1$.
So every pair solving $(3)$ gives rise to a smaller pair (whose larger member is the smaller of the original), thus creates a descending chain (and an ascending) necessarily ending in $(1,2)$, therefore it belongs to the Fibonacci family.
