How to prove that there is an infinite number of primitive pythagorean triples such as $b=a+1$ and $2 | a$?

Question

I need to prove that there is an infinite primitive pythagorean triples such as $b=a+1$ and $2 | a$ but I don't know how.

I know that $(2st, s^2-t^2, s^2+t^2)$ is a primitive pythagorean triple, then I tried to say that: $ \ 2st+1=s^2-t^2$ but it didn't work. I also tried $ \ s^2-t^2+1 = 2st$ but I don't know how to continue.

Thanks in advance

Answer 1

Only the first version should be considered because $2|a$.In that case $2t^2+1= (s-t)^2$, Replace $s-t$ by $x$ and get a Pell equation $x^2-2t^2=1$ which has infinitely many solutions.This gives you infinitely many pairs $(s, t)$ and infinitely many pythagorean triples.

Answer 2

Let Euclid's formula for generating Pythagorean triples be $F(m,k)$ where

$$A=m^2-k^2\qquad B=2mk\qquad C=m^2+k^2$$ This is a reversal of the A and B in your equations but it works the same. We can now use a formula that generates the Pell numbers needed for input to Euclid's formula which in turn generates $(B-A=\pm1)$ triples in sequential order of size. By inspection we can see that, starting with $(1)$, there are infinite solutions.

$$m=k+\sqrt{2k^2+(-1)^k}$$ For example(s) \begin{align*} k=1:\qquad & m=(1+\sqrt{2(1)^2+(-1)^1}\space)\big)=2\qquad \qquad & F(2,1)=(3,4,5)\\ k=2:\qquad & m=(2+\sqrt{2(2)^2+(-1)^2}\space)\big)=5\qquad \qquad & F(5,2)=(21,20,29)\\ k=5:\qquad & m=(5+\sqrt{2(5)^2+(-1)^5}\space)\big)=12\qquad \qquad & F(12,5)=(119,120,169)\\ k=12:\qquad & m=(12+\sqrt{2(12)^2+(-1)^{12}}\space)\big)=29\qquad \qquad & F(29,12)=(697,696,985) \end{align*}