Exercise 5.3.15* (Nearly isosceles Pythagorean triangle)

Proof. If a Pythagorean triple of integers $(x,y,z)\ne (0,0,0)$ belongs to an isosceles right triangle, then $x=y$, so $2x^2=z^2$, where $x\ne 0$, otherwise $(x,y,z)=(0,0,0)$. Therefore $\sqrt{2}=z∕x$. This is impossible, because $\sqrt{2}$ is an irrational number. This shows that no Pythagorean triple of integers belongs to an isosceles right triangle.

We prove that there are infinitely many primitive positive Pythagorean triples $(x,y,z)$ such that $y=x-1$. By Theorem 5.5, there are positive integers $r,s$ such that

So

Put $t=r-s$, so that $r=s+t$. The equation ${(r-s)}^2-2s^2=1$ becomes the Pell-Fermat equation

We have proved in Problem 10 that this equation has infinitely many solutions $(t_n,s_n)$, given by

for all $n\in \mathbb{N}$. Then $r_n=s_n+t_n$ satisfies ${(r_n-s_n)}^2-2s_n^2=1$, thus $(x_n,y_n,z_n)=(r_n^2-s_n^2,2r_n{s}_n,r_n^2+s_n^2)$ is a Pythagorean triple.

Note that $t_0=1$ is odd, and $s_0=0$ is even. If we suppose that $t_n$ is odd, and $s_n$ even, then $t_{n+1}=3t_n+4s_n$ is odd, and $s_{n+1}=2t_n+3s_n$ is even. This shows by induction that for all $n\in \mathbb{N}$,

Now we prove that $(x_n,y_n,z_n)$ is a primitive Pythagorean triple. Indeed, $t_n\wedge s_n=1$, because $t_n^2-2s_n^2=1$. Since $r_n=s_n+t_n$, we obtain $r_n\wedge s_n=1$. If $p$ is a prime divisor of $x_n$ and $z_n$, then $p\mid r_n^2-s_n^2$ and $p\mid r_n^2+s_n^2$, thus $p\mid 2r_n^2$ and $p\mid 2s_n^2$, thus $p\mid (2r_n^2)\wedge (2s_n^2)=2(r_n^2\wedge s_n^2)=2$, so $p=2$. This is impossible, because $r_n=s_n+t_n$ is odd, $s_n$ is even, therefore $p=2$ cannot divide the odd integer $r_n^2-s_n^2$. This shows that $x_n\wedge z_n=1$, a fortiori $x_n\wedge y_n\wedge z_n=1$, so $(x_n,y_n,z_n)$ is a primitive Pythagorean triple, with $x_n>0,y_n>0,z_n>0$ if $n\ge 1$, satisfying $y_n=x_n-1$.

The acute angles of the corresponding triangles are ${\alpha }_n$ and $\pi ∕2-{\alpha }_n$, where

Since $t_{n+1}>3t_n$, $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{t}_n=+\infty$, thus $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{r}_n=+\infty$. Moreover $r_n-s_n=t_n\ge 1$, thus $x_n=r_n^2-s_n^2=(r_n-s_n)(r_n+s_n)\ge r_n+s_n>r_n$, thus

Hence

if $𝜀$ is given, there exists an integer $N$ such that for every $n\ge N$, $|\pi ∕4-{\alpha }_n|<𝜀$.

So there are infinitely many primitive Pythagorean triples for which the acute angles of the corresponding triangles are, for any given positive $𝜀$, within $𝜀$ of $\pi ∕4$. □