I have a question about number of primitive Pythagorean triples.
Is there infinite number of primitive Pythagroean triples for which the acute angles of the corresponding triangles are, for any given positive $\epsilon$, within $\epsilon$ of $\pi/4$??
Thank you.
Yes. There are infinitely many where the two sides differ by 1.
$$x^2+(x+1)^2=y^2$$ $$2x^2+2x+1=y^2$$ $$(2x+1)^2+1=2y^2$$ A Pell equation, with infinitely many solutions.
3, 4, 5; 20, 21, 29; 119, 120, 169; the hypotenuse sequence satisfies $a_n=6a_{n-1}-a_{n-2}$.
As shown in this answer, the general primitive Pythagorean triple is given by $$ (m^2-n^2,2mn,m^2+n^2)\tag{1} $$ where $m\gt n$, $(m,n)=1$, and $m+n$ is odd. So the question is whether we can find $m,n$ so that $$ m^2-n^2-2mn=\pm1\tag{2} $$ Suppose we start with $$ \begin{align} m_k^2-n_k^2-2m_kn_k&=(-1)^k\tag{3a}\\ m_{k+1}m_k-n_{k+1}n_k-m_{k+1}n_k-m_kn_{k+1}&=(-1)^k\tag{3b} \end{align} $$ where $\text{(3a)}$ is true for $k\in\{-1,0\}$ and $\text{(3b)}$ is true for $k=-1$.
Furthermore, suppose we continue the sequences by $$ \begin{align} m_k&=2m_{k-1}+m_{k-2}\\ n_k&=2n_{k-1}+n_{k-2} \end{align}\tag{4} $$ Then we have, using $(3)$ and $(4)$, $$ \begin{align} &m_k^2-n_k^2-2m_kn_k\\ &=(2m_{k-1}+m_{k-2})^2-(2n_{k-1}+n_{k-2})^2-2(2m_{k-1}+m_{k-2})(2n_{k-1}+n_{k-2})\\ &=4(m_{k-1}^2-n_{k-1}^2-2m_{k-1}n_{k-1})+(m_{k-2}^2-n_{k-2}^2-2m_{k-2}n_{k-2})\\ &+\ 4(m_{k-1}m_{k-2}-n_{k-1}n_{k-2}-m_{k-1}n_{k-2}-m_{k-2}n_{k-1})\\ &=4(-1)^{k-1}+(-1)^{k-2}+4(-1)^{k-2}\\ &=(-1)^k\tag{5a} \end{align} $$ and $$ \begin{align} &m_km_{k-1}-n_kn_{k-1}-m_kn_{k-1}-m_{k-1}n_k\\ &=(2m_{k-1}+m_{k-2})m_{k-1}-(2n_{k-1}+n_{k-2})n_{k-1}\\ &-\ (2m_{k-1}+m_{k-2})n_{k-1}-m_{k-1}(2n_{k-1}+n_{k-2})\\ &=2(m_{k-1}^2-n_{k-1}^2-2m_{k-1}n_{k-1})\\ &+\ (m_{k-1}m_{k-2}-n_{k-1}n_{k-2}-m_{k-1}n_{k-2}-m_{k-2}n_{k-1})\\ &=2(-1)^{k-1}+(-1)^{k-2}\\ &=(-1)^{k-1}\tag{5b} \end{align} $$ Thus, $(3)$ holds for all $k$.
Therefore, if we start $(m_{-1},n_{-1})=(0,1)$ and $(m_0,n_0)=(1,0)$, then we get $$ \begin{align} (m_1,n_1)&=(2,1)\implies(3,4,5)\\ (m_2,n_2)&=(5,2)\implies(21,20,29)\\ (m_3,n_3)&=(12,5)\implies(119,120,169)\\ (m_4,n_4)&=(29,12)\implies(697,696,985)\\ (m_5,n_5)&=(70,29)\implies(4059,4060,5741)\\ &\vdots \end{align} $$ Using continued fraction approximations to $1+\sqrt2$, which is a root of $x^2-2x-1=0$, it can be shown that this sequence contains all of the primitive Pythagorean triples with legs that differ by $1$.
Instead of using $(4)$ to generate $m_k$ and $n_k$, we can solve the recurrence to get $$ m_{k-1}=n_k=\frac{\left(1+\sqrt2\right)^k+\left(1-\sqrt2\right)^k}{2}\tag{6} $$
