I want to find $a,b,c$ such that $ax^2 +bx+c$ is a complete cube for two values of $x$ and that quadratic equation has an integers solutions. I want to make a question for my students. Any hints? Thanks in advance
Edit: am trying to find two integer solutions for something like $(x-a)(x-b)=k^3$
Write your quadratic as $a(x-p)(x-q)$ so the roots will be $p,q$. Now you want $a(x-p)(x-q)$ to be a cube for two values of $x$. One way is to have $a$ a square and the difference between $p$ and $q$ be $\sqrt a-1$. Then $x-p$ can be $\sqrt a$ and $x-q$ can be $1$ or $x-p$ can be $-1$ and $x-q$ can be $-\sqrt a$ Do the two cubes need to be different? My construction will make them the same.
Putting this all together, we can have $a=9, p=7, q=5$, which gives the quadratic as $9x^2-108x+315$ At $x=8$ this evaluates to $27$ and at $x=4$ this is also $27$
Added: as a bonus, at $x=1$ and $x=11$ this evaluates to $216=6^3$ so you can get two different cubes from this polynomial. This has $p-q=b\sqrt a - b^2$ with $b=2 and suggests another structure for solutions.