Problem 895

The answer is no. We work modulo 4 (i.e. we look at remainders on division by 4). Note that any square is $\equiv 0$ or $1 \bmod 4$. Indeed, if $n=2 k$ is even, $n^{2}=4 k^{2} \equiv 0$, while if $n=2 k+1$ is odd, $n^{2}=4 k^{2}+4 k+1 \equiv 1$.

Now suppose some three such primes exist. If they are all odd, then each of them is $\equiv 1$ or $3(\bmod$ 4); some two must then have the same value modulo 4 (both 1 or both 3 ), and then their sum is $\equiv 2$ $(\bmod 4)$, so it cannot be a square.

Thus, the primes are not all odd, so one of them is 2 . Since they are all different, the other two are odd. If these two are both $\equiv 3$, then again their sum is $\equiv 2$ and cannot be a square. So one of them is $\equiv 1$, but then this prime plus 2 is congruent to 3 , so that sum cannot be a square. So, once again, we can find two of the three primes whose sum is not a square.