The number of primitive Pythagorean triangles with bounded hypotenuses

Question

I was reading the "mathematical constants" book. At some point, it said that Lehmer proved the following theorem in 1900.

$$\lim_{n \rightarrow \infty} \frac{P_h(n)}{n}=\frac{1}{2\pi}$$ $$\lim_{n \rightarrow \infty} \frac{P_p(n)}{n}=\frac{\ln2}{\pi^2}$$

where $P_h(n), P_p(n)$ are the number of primitive Pythagorean triangles whose hypotenuses and perimeter do not exceed $n$, respectively.

Is there any simple proof for those beautiful results?

Answer 1

Thanks to @BarryCipra, I read the original paper.

Here is the sketch of the proof without technical details:

Consider

$a=m^2+n^2$

$ b= m^2-n^2$

$c= 2mn$

It should be $m>n>0$ and $gcd(m,n)=1$ also $m,n$ has not the same parity.

If we want $hyp=a<N$ it must be $m^2+n^2<N$ so $m,n$ lies in a quarter of the circle with radius $\sqrt{N}$ and below the line $y=x$ whose area is equal to $$\frac{N\pi}{8}$$

Now the probability of $gcd(m,n)=1$ is equal to $\frac{6}{\pi^2}$ and the probability of both of $m,n$ not being odd, conditioned to be coprime is $2/3$. (This part is not so exact.)

We have $$\frac{N\pi}{8} \times \frac{6}{\pi^2} \times \frac{2}3=\frac{N}{2\pi} $$
Pythagorian triangles with hypotenuses less than $N$.