Mediterranean Mathematical Competition PETER O' HALLORAN MEMORIAL

1. For $x=3^{2n-1}$ we have: $$\begin{array}{rl}P(3^{2n-1})& =81^{2n-1}-27^{2n-1}-3\cdot (-1{)}^{2n-1}-3^{2n-1}+1\\ & \equiv 1^{2n-1}-2^{2n-1}-3\cdot 9^{2n-1}-3^{2n-1}+1(\mathrm{mod}5)\\ & \equiv 5-(2^{2n-1}+3^{2n-1})(\mathrm{mod}5).\end{array}$$ For odd exponents $m$, the number $a^m+b^m$ is divided by $a+b$. Hence $2^{2n-1}+3^{2n-1}$ is divided by $5$, and hence the number $P(3^{2n-1})$ is divided by $5$. Since $P(x)$ is a not constant polynomial it can take the value $5$ a finite number of times. Therefore there are infinite values of the polynomial divided by $5$ which are composite numbers.