Solve in positive integers the equation $a^3+b^3=9ab$

Question

Solve in positive integers the equation: $$a^3+b^3=9ab$$

I try to: $$\dfrac{a^2}{b}+\dfrac{b^2}{a}=9\Longrightarrow a^2<9b,b^2<9a$$

Of course, I can't solve it. Can anyone help?

Answer 1

By the Arithmetic Mean Geometric Mean Inequality, we have $$\frac{9}{2}=\frac{1}{2}\left(\frac{a^2}{b}+\frac{b^2}{a}\right)\ge \sqrt{ab}.$$ Now we only have a smallish number of possibilities to examine.

A simple congruential observation then dramatically cuts down the number of cases. For $a$ and $b$ must both be even. (If they are both odd, the left-hand side is even and the right-hand side is odd. If they are of different parity, the left-hand side is odd and the right-hand side is even.)

Thus $a^3+b^3$ is divisible by $8$, and one of $a$ and $b$ is divisible by $2$, and the other is divisible by $4$.

Answer 2

HINT: $x^3+y^3=9xy$ is the Cartesian equation of a Folium of Descartes $x^3+y^3=3axy$ with $a=3$. This is a rational cubic with a double point and its parametric equations are $x=\frac{9t}{1+t^3}$ and $y=tx$ The loop of the curve occurs for the values $t\gt 0$ so the integer points we are looking for belong to this loop.

For $t\ge 3$ one has $x\lt 1$ hence possible integers values $x, y$ correspond to values of the parameter $t$ such that $0\lt t \lt 3$.

When $t=1$ there is not integer solution $(a,b)$ but when $t=2$ there is the solution $\color {red}{(a,b)=(2,4)}$ (and trivially $(b,a)$ from symmetry over $y=x$).

Now,to discard fractional values of $t$, taking derivatives, we have $$x’=\frac{9-18t^3}{(1+t^3)^2}$$ $$y’=\frac{18t-9t^4}{(1+t^3)^2}$$ it follows $$\max x=\frac{9\sqrt[3]{2}}{1+2}=3\sqrt[3]{2}\approx 3.7797$$ $$\max y=\frac{9\sqrt[3]{4}}{1+2}=3\sqrt[3]{4}\approx 4.7622$$

Consequently there are just $12$ points to verify (in the “lattice” $\{1,2,3\}\text {x}\space \{1,2,3,4\}$). Thus the only solution in positive integers is the given one $(2,4)$.