As claimed by Fermat and proved by Euler, any prime $p\equiv1\pmod4$ can be written uniquely as $s_p^2+t_p^2$ with $s_p,t_p\in\{1,2,3,\ldots\}$, $2\nmid s_p$ and $2\mid t_p$. For any positive integer $n$, let us define $$S(n):=\sum_{p\le n\atop p\equiv1\pmod4}s_p \ \ \ \text{and}\ \ \ T(n)=\sum_{p\le n\atop p\equiv1\pmod4}t_p.$$ Via computation, I have found that $$S(10^9)=334976550299,\ \ T(10^9)=334979004134,\ \ \frac{S(10^9)}{T(10^9)}\approx 0.99999267.$$ This leads me to pose the following conjecture.
Conjecture. We have $$\lim_{n\to+\infty}\frac{S(n)}{T(n)}=1.$$
QUESTION 1. Is the conjecture true? If true, how to prove it?
I have another question.
QUESTION 2. Is there a positive contant $c$ such that $$\lim_{n\to+\infty}\frac{\sum_{p\le n\atop p\equiv1\pmod4}s_t/t_p}{\sum_{p\le n\atop p\equiv1\pmod4}t_p/s_p}=c$$ holds?
Concerning this question, I conjecture that $c$ exists and itits value is approximately 0.87probably $1$. I have found that $$\frac{\sum_{p\le 10^9\atop p\equiv1\pmod4}s_t/t_p}{\sum_{p\le 10^9\atop p\equiv1\pmod4}t_p/s_p}\approx 0.87085958.$$$$\frac{\sum_{p\le 10^{11}\atop p\equiv1\pmod4}s_t/t_p}{\sum_{p\le 10^{11}\atop p\equiv1\pmod4}t_p/s_p}\approx 0.896.$$
Your comments are welcome!