\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION

Let $f(x)=x^3-x+1$

Calculate the values of $f(0),\ f(1),\ f(2)$ modulo 3, and hence
deduce that $f(x)=0$ has no integer solutions.



ANSWER

$f(0)=0-0+1=1,\ f(1)=1-1+1=1$ and $f(2)=8-2+1=7$, and so $f(0),
f(1)$ and $f(2)$ are all congruent to 1 modulo 3. If $n$ were an
integer solution of $f(x)=0$, we'd have $f(n)=0$, and hence
$f(n)\equiv0$ mod 3. But $n$ must be congruent to one of 0,1,2 mod
3, by lemma 3.3, and so by the above this gives $f(n)\equiv1$ mod
3. This contradiction shows that we cannot have any integer
solutions of $f(x)=0$.




\end{document}