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

Question

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?

Answer 1

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.

Answer 2

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.

So you're looking for a formula involving the cube root of a square root of a negative number involving a series of squares, cubes and other powers, and hoping it will come out to be an integer.

At this point I decided to stop attempting this problem, since it looks like there is not much for it except to search for candidates using a computer, and I don't want to do that right now. However, I decided to post this partial answer since it might save someone some work. I'm sorry that I couldn't give a better answer, but I hope this at least moves things in the right direction.