Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 4 / 1E (Numbers and Sets)

The Euclidean algorithm givesBack-substitution supplies the Bezout identityThus one solution is $(x,y)=(-5,2)$. If $(x,y)$ is any solution, subtraction gives $7(x+5)+18(y-2)=0$. Since $7$ and $18$ are coprime, $18$ divides $x+5$. Write $x+5=18t$; substitution then gives $y-2=-7t$. Conversely these values satisfy the equation for every integer $t$. Hence all solutions are

Factor the expression as $n^3-n=(n-1)n(n+1)$. One of three consecutive integers is divisible by $3$, so $3\mid n^3-n$. If $n$ is odd, its neighbours $n-1$ and $n+1$ are consecutive even integers. One is divisible by $4$ and the other by $2$, so their product is divisible by $8$. Since $3$ and $8$ are coprime integers, their product divides the expression:The argument also includes negative odd integers and the zero product at $n=\pm 1$.