Let $f \in \Bbb Z[x]$ such that there exist distinct $a$, $b$, $c$ and $f(a)=f(b)=f(c)=1$. Prove that there is no $d \in \Bbb Z$ such that $f(d)=0$

Question

I'm having trouble trying to solve this problem.

Let $f \in \Bbb Z[x]$ such that there exist $a$, $b$, $c$ (all different) and $$f(a)=f(b)=f(c)=1.$$ Prove that there is no $d \in \Bbb Z$ such that $f(d)=0$.

I'm practicing for an exam.

I've tried setting a polynomial $g(x)=f(x)-1$ and factoring it as if it had three roots, but then I do not know how to continue.

Any suggestions?

Should it be $f \in \mathbb{Z}[x]$? – Matt Apr 23 '17 at 22:00 — Matt, Apr 23 '17 at 22:00

score 18 · Accepted Answer · answered Apr 23 '17 at 22:07

You were in the right path, the idea is that if you have three distinct numbers in $\mathbb{Z}$ one must be different from $\pm1$.

Take $g(x)$ as you have defined and assume there is $d$ such that $f(d)=0$, then

$$(d-a)(d-b)(d-c)\tilde{g}(d)=-1$$

A contradiction since $d-a,d-b,d-c$ are distinct.

score 9 · Answer 2 · answered Apr 23 '17 at 22:09

Your start is good.

So, we can write $f(x) =u(x)\cdot(x-a)(x-b)(x-c) + 1$ with some polynomial $u\in\Bbb Z[x]$.

Then, for any distinct integer $d$, we have that at least one of $|d-a|,\ |d-b|,\ |d-c|$ is $\ge 2$, and thus $u(d)\cdot(d-a)(d-b)(d-c)$ can't ever be $-1$.

Let $f \in \Bbb Z[x]$ such that there exist distinct $a$, $b$, $c$ and $f(a)=f(b)=f(c)=1$. Prove that there is no $d \in \Bbb Z$ such that $f(d)=0$

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