Solutions to the Diophantine equation $(ab)(a+b) = N$

Question

I was working on a problem that I already posted to SE that dealt with the problem $(ab)(a + b) = 240$ and I was curious to look for solutions to a general $N$ instead of just $240$. In simple terms, I asked myself: what numbers can be written as the sum of two numbers times their product?

After some basic trial and error work, I found that the first few numbers that met these criteria were, $0, 2, 6, 12, 20, 30\dots$.

Example. $(4\cdot 1)(4 + 1) = 5\cdot4 = 20$.

Clearly, odd numbers will never appear in this list as at least one of the sum or product of two numbers will always be even.

I then found a clear method to generate these numbers from all base pairs

\begin{matrix} (1, 1) & (1, 2) & (1,3)\\ & (2, 2) & (2, 3) \end{matrix}

which yielded the complete list: $$2, 6, 12, 16, 20, 30, 42, 48, 54, 56, \dots$$

that I found on OEIS.

I also noticed that some values can be written in two distinct ways:

E.g. $30 = (5\cdot1)(5+1) = (3\cdot2)(3+2)$

which closely resembles the study of taxicab numbers. I am wondering, has this sequence/number type been studied very much? I am thinking of looking into this topic for a undergraduate thesis in elliptic curves, and this looks like an approachable topic (yet niche enough for possible new results).

Answer 1

$ab(a+b)=N$

Above has parametric solution:

$a=8w$

$b=3w$

$N=264w^3$

For, $w=7$, we have

$(a,b,c)=(56,21,90552)$

Answer 2

You set the problem up as a quadratic equation.

Wlog render $a\ge b$. Then $2b^3\le N$ and $b\le\sqrt[3]{N/2}$. Now for any given $b$ that meets this bound and divides $N$ we may set the equality up as a quadratic equation for $a$; thus

$(b)a^2+(b^2)a-N=0, \Delta=b^4+4bN$

If the discriminant $\Delta$ is a perfect square, then the quadratic equation may be solved for a positive whole number root $a$ which, paired with the value of $b$ that gives this correct discriminant, will solve $ab(a+b)=N$.

Let's pick $N=750$ for fun. Then the bound on $b$ is $\sqrt[3]{375}<8$ and the divisors of $750$ within this bound are

$1,2,3,5,6.$

So

$b=1\implies\Delta=3001$ $b=2\implies\Delta=6016$ $b=3\implies\Delta=9081$ $b=5\implies\Delta=15625=\color{blue}{75^2}$ $b=6\implies\Delta=19296$

Of these possibilities, only $b=5$ gives a square value for $\Delta$ (the other discriminants are quadratic nonresidues $\bmod 7$, $\bmod 7$, $\bmod 11$ and $\bmod 11$ respectively), and then the quadratic equation for $a$ is

$5a^2+25a-750=0\implies a^2+5a-150=0\implies a=10(a>0)$

$\therefore (10×5)(10+5)=750.$