TITLE: $\sum p\;$ over consecutive primes $\;p \equiv 3 \pmod 4\;$ is never a square?
QUESTION [6 upvotes]: Prove or disprove that $\;\sum p\;$ over consecutive primes $\;p \equiv 3 \pmod 4$, starting from $\,p=3$, is never a square.
I think it would be useful to examine the modular equation $\;\sum p \equiv x^2 \pmod 4$.
Many thanks for your help.

REPLY [3 votes]: I conjecture that the sum is a square (rarely but) infinitely often. (Moreover, this has nothing to do with the precise sequence being summed—the conjecture would be the same for any natural sequence growing at the same rate.)
The sum of the first $n$ primes that are congruent to $3$ (mod $4$) is asymptotic to $n^2\log n$. When adding the next such prime, which is about $2n\log n$ in size, we will jump over approximately $\sqrt{\log n}$ squares, which means that the "probability" of landing on a square should be the same as the probability that a randomly chosen number of size about $n^2\log n$ is a square, which is $\frac1{n\sqrt{\log n}}$. Since $\sum \frac1{n\sqrt{\log n}}$ diverges (at a rate of about $2\sqrt{\log n}$, by the integral test), this suggests that there should be infinitely many examples.