Berkeley Math Circle

Solution:

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 $n=2k$ is even, $n^{2}=4k^{2} \equiv 0$, while if $n=2k+1$ is odd, $n^{2}=4k^{2}+4k+1 \equiv 1$.

Now suppose some three such primes exist. If they are all odd, then each of them is $\equiv 1$ or $3 \pmod{4}$; some two must then have the same value modulo $4$ (both $1$ or both $3$), and then their sum is $\equiv 2 \pmod{4}$, 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.