THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

a) We have $a_n=a_1+(n-1)r$, $n\in {\mathbb{N}}^{\ast }$, with $a_1\in {\mathbb{N}}^{\ast }$, $r\in \mathbb{N}$. This yields $$S_p=p\left(a_1^2+a_1(p-1)r+\frac{(p-1)(2p-1)}{6}{r}^2\right).$$ Since $p\ge 5$ is a prime number, we have $\gcd (p,2)=1$, $\gcd (p,3)=1$, hence $6\mid (p-1)(2p-1)$, that is, $p\mid S_p$.

b) Assume, by way of contradiction, that $S_5$ is a square. From a), we deduce that for some $y\in {\mathbb{N}}^{\ast }$, we have $S_5=(5y{)}^2$. Writing $S_5$ in terms of $x=a_3\in {\mathbb{N}}^{\ast }$ and the common difference $r$ we obtain $S_5=5x^2+10r^2$, hence the equation $x^2+2r^2=5y^2$. It is not difficult to see that the equation has no solution in positive integers, by checking it modulo $5$.