Suppose $p, q$ ∈ N are prime numbers so that $x^3 −px+q$ has only integer roots. Solve for p and q

Question

I've got no idea how to do this question. Any help would be appreciated

Answer 1

By Vieta's formulae: $$x_1x_2x_3=-q$$ Where $q$ is prime, $x_1,x_2,x_3\in\mathbb Z$

Let $x_1x_2=-1$ and $x_3=q$ or $x_1x_2=1$ and $x_3=-q$ We obtain: $$x_1=-1,x_2=1,x_3=q$$ or $$x_1=1,x_2=1,x_3=-q$$

Then, for $x=1$: $$1-p+q=0\;\land\; p\;\text{is prime}\implies p=q+1=2+1=3$$ Because there are only 2 consecutive prime numbers: $2, 3$

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Edit:

as written above, one root definitely equals $1$, so: $(x-1)\mid x^3-3x+2$

Long division: $$(x^3-3x+2):(x-1)=x^2+x-2=x^2-x+2=(x-1)(x+2)\implies x^3-3x+2=(x-1)^2(x+2)$$ and we see the roots are our $2^{\text{nd}}$ option.

Answer 2

If $q$ is a root, so $$q^3-pq+q=0,$$ which gives $$q^2+1=p,$$ which is possible for $q=2$ and $p=5$ only.

Thus, $$x^3-5x+2=0$$ or $$x^3-2x^2+2x^2-4x-x+2=0$$ or $$4(x-2)(x^2+2x-1)=0,$$ which is a contradiction.

If $-q$ is a root, we obtain $$-q^3+pq+q=0,$$ which gives $$p=q^2-1,$$ which is possible for $q=2$ and $p=3$ only, which gives $$x^3-3x+2=0$$ or $$(x-1)^2(x+2)=0,$$ which is valid.

We have no another cases.