Leo Giugiuc posted an elegant problem and its solution at the CutTheKnotMath facebook page.

Let $P(x)=x^2+ax+b,$ be a monic (with the leading coefficient $!)$ quadratic polynomial, with integer $a$ and $b.$ Prove that, for any integer $n,$ there is an integer $m$ such that $P(n)P(n+1)=P(m).$

### Proof

Let $u$ and $v$ be the roots of $P(x)=0.$ It does not matter whether they are integers or not. $P(x)=(x-u)(x-v)$ and $P(x+1)=(x+1-u)(x+1-v),$ implying

$P(x)P(x+1)=(x-u)(x-v)(x+1-u)(x+1-v).$

Now,

\begin{align} (x-u)(x+1-v) &= x^{2}+x-xv-xu-u+uv\\ &=x^{2}-x(u+v)+uv+x-u\\ &=x^{2}+x(a+1)+b-u. \end{align}

Similarly, $(x-v)(x+1-u)=x^{2}+x(a+1)+b-v,$ meaning that

\begin{align} P(x)P(x+1) &= (x^{2}+x(a+1)+b-u)(x^{2}+x(a+1)+b-v)\\ &=P(x^{2}+x(a+1)+b). \end{align}

Thus, for a given $n,$ $m=n^{2}+n(a+1)+b,$ an integer.

Nothing short of brilliant!