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

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.

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

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