# A mysterious connection between primes and $\pi$

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!

• 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? – Zhi-Wei Sun Dec 16 '19 at 11:18
• 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. – GH from MO Dec 16 '19 at 11:28
• I conjecture further that $$\frac{s_*(n)}{s(n)}=\frac1{\pi}+O\left(\frac1{\sqrt n}\right).$$ – Zhi-Wei Sun Dec 16 '19 at 15:50
• 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. – GH from MO Dec 16 '19 at 20:19
• 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/… – Nilotpal Kanti Sinha Dec 19 '19 at 12:10

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 $$\lim_{n\to\infty}\frac{s_\ast(n)}{s(n)}=\frac{1}{\pi}.$$
• 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 $$\lim_{n\to+\infty}\frac{s(n)}{s_*(n)} = 1?$$ – Mike Dec 17 '19 at 11:50
• @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. – GH from MO Dec 17 '19 at 12:47
• 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. – Mike Dec 18 '19 at 9:04
• @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. – GH from MO Dec 18 '19 at 14:11