Find integer solutions to $x^2y + xy^2 = 240$

Question

This question is an exercise in chapter $5$ of Tate & Silverman's Rational Points on Elliptic curves.

Find all integer solutions to $x^2y + xy^2 = 240$.

Here is my work:

First we factor to find that
\begin{align*}
x^2y + y^2x &= 240 \\
xy(x + y) &= 240.
\end{align*}

Now we list all possible factor pairs of $240$:

$$240\cdot1, 120\cdot 2, 4\cdot 60, 8\cdot 30, 16 \cdot 15, 3\cdot80, 6\cdot40, 12\cdot 20, 24\cdot 10, 48\cdot 5$$

We are looking for pairs in which one number can be written as $xy$ and another as $x + y$. By inspection, we find that the pairs $16\cdot 15, 12\cdot 20$ and $24\cdot 10$ are the only ones that work. So we have that

\begin{align*}
16 \cdot 15 &\implies x = 15,\: y = 1 \\
12 \cdot 20 &\implies x = 10,\: y = 2 \\
24 \cdot 10 &\implies x = 6,\: y = 4
\end{align*}
which yields the solution set $\{(15, 1),\: (1, 15),\: (10, 2),\: (2, 10),\: (6, 4),\: (4, 6)\}$.

Is this method and solution set correct?

Answer 1

The possible solutions of the Diophantine equation $xy(x+y)$ that I got from your method are (assuming negative integer values are allowed): $(-10,4),(-10,6),(1,15),(1,-16),(15,1),(-16,1),(2,10),(2,-12),(4,6),(4,-10),(6,4),(6,-10), (10,2),(-12,2)$.