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
$$\sum_{j=1}^{42} \frac{1}{p_j^2 + 1} = \frac{1}{n^2} \iff n^2 \cdot \sum_{j=1}^{42} \frac{P}{p_j^2 + 1} = P.$$
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 $P \equiv 1 \pmod{3}$ 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 $3^2 + 1 \equiv 1 \pmod{3}$, and therefore $P \equiv -1 \pmod{3}$. 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 $n^2 \equiv 0 \pmod{3}$ or $n^2 \equiv 1 \pmod{3}$, 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