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The Prime Number Theorem relates primes to the important constant $e$.

Here I report my following surprising discovery which relates primes to $\pi$.

Conjecture (December 15, 2019). Let $s(n)$ be the sum of all primes $p\le n$ with $p\equiv1\pmod4$, and let $s_*(n)$ be the sum of those $x_py_p$ with $p\le n$, where $p$ is a prime congruent to $1$ modulo $4$, and $p=x_p^2+y_p^2$ with $x_p,y_p\in\{1,2,3,\ldots\}$ and $x_p\le y_p$. Then $$\lim_{n\to+\infty}\frac{s(n)}{s_*(n)} = \pi.$$

Recall that a classical theorem of Euler (conjectured by Fermat) states that any prime $p\equiv1\pmod4$ can be written uniquely as $x^2 + y^2$ with $x,y\in\{1,2,3,\ldots\}$ and $x\le y$. Since $x^2 + y^2 \ge 2xy$ for any real numbers $x$ and $y$, we have $s(n) \ge 2s_*(n)$ for all $n=1,2,3,\ldots$.

I have created the sequence $(s_*(n))_{n>0}$ for OEIS (cf. http://oeis.org/A330487). Via computation I found that $$s(10^{10}) = 1110397615780409147,\ \ s_*(10^{10}) = 353452066546904620, $$ and $$ 3.14157907 < \frac{s(10^{10})}{s_*(10^{10})} < 3.14157908. $$ This looks an evidence to support the conjecture.

QUESTION. Is my above conjecture true? If true, how to prove it?

Any further check of the conjecture is welcome!

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    $\begingroup$ Dear GH from MO, thank you for your clever answer. I cannot even find your e-mail address. Would you please send me an e-mail so that we may discuss more on such topics? $\endgroup$
    – Zhi-Wei Sun
    Commented Dec 16, 2019 at 11:18
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    $\begingroup$ Thanks for your kind words and the nice conjecture. I prefer to remain anonymous and pursue discussions at this site. If I see an interesting question, and I have the ability and time to answer it, I will answer it. $\endgroup$
    – GH from MO
    Commented Dec 16, 2019 at 11:28
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    $\begingroup$ I conjecture further that $$\frac{s_*(n)}{s(n)}=\frac1{\pi}+O\left(\frac1{\sqrt n}\right).$$ $\endgroup$
    – Zhi-Wei Sun
    Commented Dec 16, 2019 at 15:50
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    $\begingroup$ I think that an error term $O(n^{-1/2})$ is too ambitious, the truth is probably $O(f(n)n^{-1/2})$ and $\Omega_{\pm}(f(n)n^{-1/2})$ with some slowly increasing $f(n)\to\infty$. But to prove that one probably needs the Riemann Hypothesis for the relevant Hecke $L$-functions. By analogy, a good error term for $\pi(x;q,a)/\pi(x)$ is hard to obtain. $\endgroup$
    – GH from MO
    Commented Dec 16, 2019 at 20:19
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    $\begingroup$ Motivated by this question, I found a related conjecture. The ratio of the sum of the squares of the hypotenuse to the sum of the area of all Pythagorean triangles in which the hypotenuse is a prime number is $2\pi$. Posted in MSE: math.stackexchange.com/questions/3481766/… $\endgroup$
    – Nilotpal Kanti Sinha
    Commented Dec 19, 2019 at 12:10

1 Answer 1

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Here is a proof of the conjecture. We shall use Hecke's theorem that the angles of the lattice points $(x_p,y_p)$ are asymptotically equidistributed in $[\pi/4,\pi/2]$, cf. this MO post.

Let $t_p\in[\pi/4,\pi/2]$ be the angle of the lattice point $(x_p,y_p)$. Let us divide the interval $[\pi/4,\pi/2]$ into $R$ subintervals of equal length, where $R$ is large but fixed. For $r\in\{1,\dotsc,R\}$, the $r$-th subinterval is $$I_r:=[u_{r-1},u_r]\qquad\text{with}\qquad u_r:=\frac{\pi}{4}\left(1+\frac{r}{R}\right).$$ Observe that $$\frac{\sin(2u_r)}{2}\sum_{\substack{p\leq n\\t_p\in I_r}}p\leq \sum_{\substack{p\leq n\\t_p\in I_r}}x_p y_p\leq \frac{\sin(2u_{r-1})}{2}\sum_{\substack{p\leq n\\t_p\in I_r}}p.$$ By the quoted equidistribution theorem, $$\sum_{\substack{p\leq n\\t_p\in I_r}}p\sim\frac{s(n)}{R}\qquad\text{as}\qquad n\to\infty,$$ and so we infer that $$\frac{1}{R}\sum_{r=1}^R\frac{\sin(2u_r)}{2}\leq \liminf_{n\to\infty}\frac{s_\ast(n)}{s(n)}\leq \limsup_{n\to\infty}\frac{s_\ast(n)}{s(n)}\leq \frac{1}{R}\sum_{r=1}^R\frac{\sin(2u_{r-1})}{2}.$$ By letting $R\to\infty$, both sides tend to $$\frac{4}{\pi}\int_{\pi/4}^{\pi/2}\frac{\sin(2u)}{2}\,du=\frac{1}{\pi},$$ whence $$\lim_{n\to\infty}\frac{s_\ast(n)}{s(n)}=\frac{1}{\pi}.$$

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  • $\begingroup$ I check those p is the sum of three squares, such as $3=1^2 + 1^2 + 1^2,29=2^2 + 3^2 + 4^2$, it seems $$\lim_{n\to+\infty}\frac{s(n)}{s_*(n)} = 1?$$ $\endgroup$
    – Mike
    Commented Dec 17, 2019 at 11:50
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    $\begingroup$ @Mike: If $p\equiv 3\pmod{8}$, then $p$ is a some of three squares but not a sum of two squares. For such primes $p$, the quantities $x_p$ and $y_p$ are not defined. $\endgroup$
    – GH from MO
    Commented Dec 17, 2019 at 12:47
  • $\begingroup$ I check the data time again, those $p$ not include primes also can be the sum of two squares, and $x_py_pz_p$ is if $x_p+y_p+z_p$ is the min one. $\frac{s(n)}{s_*(n)}$ is close to $1$ when $n$ go large, I don't know if the value will less than $1$ when $n$ is sufficiently large. If only check primes is the sum of three consecutive squares,I believe there's a constant. $\endgroup$
    – Mike
    Commented Dec 18, 2019 at 9:04
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    $\begingroup$ @Mike: I think you did not understand my point.The definition of $s(n)$ and $s^\ast(n)$ only involves primes that can be written as a sum of two squares. If you want to talk about an analogous question involving primes which can be written as a sum of three squares, then you need to modify the definition of $s(n)$ and $s^\ast(n)$ first. So I suggest that you open a new question, starting with precise (new) definitions tailored for your situation. $\endgroup$
    – GH from MO
    Commented Dec 18, 2019 at 14:11

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