Let $f(x)$ be a polynomial with integer co-efficients. If $f(x)=2$ for three distinct integers, then for no integer $x$, can $f(x)$ be $3$. Prove.

Let $a,b,c$ be three distinct integers such that $f(a)=f(b)=f(c)=2$. Let $x$ be any integer, then if $f(x)=1+1+1$, for some $x$, then $f(x)=\frac{f(a)+f(b)+f(c)}{2}$. But how do we conclude that no such $x$ exists? How do we use the Pigeonhole Principle in this case? What are the pigeonholes and what are pigeons?


1 Answer 1


Say the three integers are $a,b,c$. Then $$p(x)=f(x)-2=(x-a)(x-b)(x-c)q(x)$$

where $q(x)$ is again a polynomial with integer coefficients. We seek to prove that $p(n)$ can never be $1$ (for integer $n$).

Suppose, to the contrary, that we had such an $n$. Then $$(n-a)(n-b)(n-c)q(n)=1$$

But $n-a,n-b,n-c$ are all integers and at least one of them is $\neq \pm 1$. That integer would be a divisor of $1$, an absurdity.

  • $\begingroup$ Thank you. A very crisp and beautiful solution... $\endgroup$
    – JackT
    Nov 24, 2017 at 13:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .