# Integer solutions of the cubic equation $x^3-a^2bx-1-2ab=0$

Given the equation $$x^3-a^2bx-1-2ab=0$$. Is there a way to know if any integer solutions exist for $$a,b$$ integers greater than 1.

I've plotted graphs and tried to brute force it but found no solutions.

Update: If you set a = 1 we get a pretty simple equation: $$x^3-bx-1-2b$$. It seems like there would be a clear way to show that no integer solution exists for $$b > 2$$. Is there any restraints that would make it easier, for example letting $$b,x$$ be prime?

• According to the rational root theorem, the only possible integer roots are factors of $-1-2ab$. Oct 20, 2022 at 2:22

Brute force search reveals that $$a=7,b=6$$ give $$0=x^3-a^2bx-1-2ab=x^3-294x-85=(x+17)(x^2-17x-5).$$ Since you only specified $$a,b$$ to be positive integers and $$x$$ be any integer, $$x=-17$$ gives a valid solution. My search did not yield any other for $$2 \leq a,b \leq 1000$$ however.
Putting $$p=-a^2b$$ and $$q=-1-2ab$$ into the depressed cubic formula, we can obtain a solution by substituting $$x=z-{p\over3z}$$ which turns the depressed cubic into a quadratic in $$z^3$$. Then the solution of the quadratic is $$z^3=-{q\over2}+\sqrt R.$$ where $$R$$ is the discriminant of the quadratic, the "$$b^2-4ac$$" part of the problem.
The formula for the discriminant, $$R$$, is \begin{align} R&=\left({p\over3}\right)^3+\left({q\over2}\right)^2\\ &=-{a^6b^3\over27}+{4a^2b^2+4ab+1\over 4} \end{align} For $$a$$, $$b$$ integers greater than one this is almost always negative, since it is negative if $$a>2$$ or $$b>2$$, so you have complex conjugate roots. If $$a=b=2$$ then the original equation is $$x^3-8x-9=0$$, which doesn't have any integer roots, since by the rational root theorem the roots must be $$\pm1,3$$, and these are not roots.
• I'm a little confused on how you get to the part $z^3 = -q/2 + \sqrt{R}$ Oct 20, 2022 at 4:48