Cool solution

While the answer is well-known, I happened to come up with this solution myself a while back and it was pretty cool, so I've gotten around to creating a problem for it.

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According to Fermat's Christmas theorem, the sum must run through all primes congruent to 1 modulo 4.

Dirichlet's theorem on primes in an arithmetic progression states that $\displaystyle \sum_{p \equiv 1 \pmod{4}} \frac{1}{p}$ diverges.

The sum in question contains the above sum, so therefore it too must diverge.

Consider the annulus $n-1 < |z| \le n$ for all $n \in \mathbb{Z}^+$ . For each $x \in \{0,1,2,\ldots,n-1\}$ , there exists $y \in \mathbb{Z}$ such that $x + iy$ is in the annulus. (Proving it is easy; when $|x+iy_1| = n-1$ , we have $y_1 = \sqrt{(n-1)^2 - x^2}$ , and when $|x+iy_2| = n$ , we have $y_2 = \sqrt{n^2 - x^2}$ . It can be verified that $y_2 - y_1 \ge 1$ , thus there is an integer $y$ such that $y_1 < y \le y_2$ . Then $|x+iy_1| < |x+iy| \le |x+iy_2|$ , so $x+iy$ is in the annulus.) For this $x+iy$ , we have $|x+iy| \le n$ , thus $\frac{1}{|x+iy|^2} \ge \frac{1}{n^2}$ . Each annulus contributes at least $n$ points, so each annulus contributes at least $n \cdot \frac{1}{n^2} = \frac{1}{n}$ to the sum. Summing over all annuli gives the harmonic series $\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \ldots$ which diverges; the given sum is not less than this number, so it also diverges.