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Find the roots of all monic cubics $f(x)$ given $f(2)=1$ and all roots are integral

I start with $f(x)=x^3-ax^2+bx-c$, $$8-4a+2b-c=1$$ $$-4a+2b-c=-7$$

At this point it seems really difficult like I do not have enough information (I do) and I'm not sure if I've already taken a bad route or if there is a good move from here.

How can I progress with this question? The answer is to be given in 3-tuples of roots.

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  • $\begingroup$ Can you find an integer which is not a root of such an equation? $\endgroup$
    – markvs
    Aug 30, 2020 at 6:08
  • $\begingroup$ @JCAA No, but there are finitely many tuples (and a small number) where all roots are integers, per the question. $\endgroup$
    – jc5535
    Aug 30, 2020 at 6:12

1 Answer 1

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Since $f$ is monic and all roots are integral, we have $$f(x)=(x-p)(x-q)(x-r)$$ for some integers $p,q,r$. Then $$(2-p)(2-q)(2-r)=f(2)=1$$ Since $(2-p),(2-q),(2-r)\in\mathbb{Z}$, there are only few possibilities for $p,q,r$, namely $p,q,r\in\{1,3\}$.

Without loss of generality, let $p\leq q\leq r$. Product of three integers is $1$ only when either (1) all of them are $1$ or (2) one of them is $1$ and the other two are $-1$. Hence we have the following two cases respectively,

$$p=q=r=1\tag{1}$$ $$p=1, q=r=3\tag{2}$$

Hence there are only two possible monic cubic polynomials

$$f(x)=(x-1)^3\tag{$1'$}$$ $$f(x)=(x-1)(x-3)^2\tag{$2'$}$$

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