Integral solutions to $a^3x^3+a^2x^2+ax+a=0$.

Question

I have to determine every possible nonzero $a\in\Bbb{Z}$ for which the equation below has an integral solution $x\in\Bbb{Z}$. I guess it is a functional equation.

This is the equation:

$$a^3x^3 + a^2x^2 + ax + a = 0$$

I have done this so far:

$$a(a^2x^3 + ax^2 + x + 1) =0 $$

$$a^2x^3 + ax^2 + x + 1 = 0$$

If I follow the comment below the root rational theorem says that $a_0$ is divisible by $p$ and $a_n$ is divisible by $q$

So I end up with:

$a^2 = p$

$1 = q$

Because:

$x_0=\tfrac{p}{q}$

while:

$p,q \in\Bbb{Z}$

Then:

$x_0 =\tfrac{a^2}{1}$

All whole numerical divisors of

$\tfrac{a^2}{1}$

which is the same as $a^2$,

are possible solutions. I still don't get how to continue.

Answer 1

You are on the right track, but you have misapplied the rational root theorem; if $$a^2x^3+ax^2+x+1=0,$$ where $a$ is an integer and $x$ is rational, then if $x=\frac{p}{q}$ is in lowest terms then $p$ divides $1$ and $q$ divides $a^2$. This means $p=\pm1$ and $x$ is an integer if and only if also $q=\pm1$, in which case $x=\pm1$. Now the question remains; for which values of $a$ does the equation have a root $\pm1$?

Answer 2

Hint Reduce the derived equation $$a^2 x^3 + a x^2 + x + 1 = 0$$ modulo $x$. (NB $x = 0$ never gives a solution.)

This leaves $$1 \equiv 0 \pmod x,$$ so $x = \pm 1$. Substitute both of these possibilities into the derived equation and check whether the resulting equation has any (nonzero) integer solutions $a$.