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I understand that "a postive integer $n$ is representable as the sum of two squares, $n=x^2+y^2$ if and only if every prime divisor $p\equiv 3$ mod $4$ of $n$ occurs with even exponent", as mentioned in this answer, so I was wondering what is the proportion of positive integers that can be expressed as the sum of two squares.

The probability that a positive integer has a prime divisor $p\equiv3\pmod4$ with an even exponent is $$1-\frac1p+\frac1{p^2}-\frac1{p^3}+\cdots=\frac1{1+\frac1p}=\frac{p}{p+1}$$by PIE, so the probability that a positive integer can be expressed as the sum of two squares would be the product of $\frac{p}{p+1}$ for all primes $p\equiv3\pmod4$, so $$\prod_{\text{prime }p\equiv3\pmod4}^\infty\frac{p}{p+1}$$

After having a computer program list this product for all the primes $p\equiv3\pmod4$ up to $100,000$, I cannot tell if this product will converge. My question is, does this infinite product converge, and is there a special closed form for this value?

Sai Mehta
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    It diverges to $0$ (this is a standard convention for infinite products, to be consistent with what happens when you take their logarithms). This is a special case of Dirichlet's theorem. – Qiaochu Yuan Jul 22 '24 at 18:26
  • See art of problem solving. One really says that it diverges to zero. Concerning your product, see this post. – Dietrich Burde Jul 22 '24 at 18:53
  • A more precise version is the last chapter (of volume 2) in Topics in Number Theory by William J. LeVeque. The count of numbers up to positive $x$ is $\frac{bx}{\sqrt{\log x}}$ asymptotically, where constant $b \approx 0.7642$ – Will Jagy Jul 23 '24 at 01:55

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