AustriaMO2011

We assume that the sum in question can be written as the reciprocal of a perfect square $n^2$. Let $P:={\prod }_{j=1}^{42}(p_j^2+1)$ be the product of all denominators of the summed fractions. We then have $$\sum _{j=1}^{42}\frac{1}{p_j^2+1}=\frac{1}{n^2}⟺n^2\cdot \sum _{j=1}^{42}\frac{P}{p_j^2+1}=P.$$ We now consider both sides of this equation modulo 3.

Case 1: If none of the numbers $p_j$ is equal to 3, each factor $p_j^2+1$ is congruent to $-1$ modulo 3. We therefore have $P\equiv 1(\mathrm{mod}3)$ and each expression $\frac{P}{p_j^2+1}$ is congruent to $-1$ modulo 3. The left side of the equation is therefore divisible by 3, which yields a contradiction.

Case 2: If $p_j=3$ holds for some index $j$, we have $3^2+1\equiv 1(\mathrm{mod}3)$, and therefore $P\equiv -1(\mathrm{mod}3)$. In the sum ${\sum }_{j=1}^{42}\frac{P}{p_j^2+1}$ we therefore have one number congruent to $-1$ (mod 3) and 41 congruent to 1. The sum is therefore congruent to 1, and since we either have $n^2\equiv 0(\mathrm{mod}3)$ or $n^2\equiv 1(\mathrm{mod}3)$, the left side is certainly not congruent to $-1$, which again yields a contradiction.

We see that the sum in question cannot be the reciprocal of a perfect square, as claimed. qed