Generating Pythagorean Triples S.T. $b = a+1$ I am looking for a method to generate Pythagorean Triples $(a,b,c)$. There are many methods listed on Wikipedia but I have a unique constraint that I can't seem to integrate into any of the listed methods.
I need to generate Pythagorean Triples $(a,b,c)$ such that:
$$a^2 + b^2 = c^2$$
$$a\lt b\lt c \,; \quad a,b,c \in \Bbb Z^+$$
$$and $$
$$b=a+1$$
Is there a way to modify one of the listed methods to include this constraint?

To produce Pythagorean triples where $a^2 + b^2 = c^2$ and $b = a + 1$
Start with $n = 1$
$\sqrt{n(1 + \sqrt{2})^4}$
Round this down to an integer, then square it.  This gets you $c^2$, which is $25$, so $c = 5$. Now take $c$ and divide it by $\sqrt{2}$.  $\dfrac{c}{\sqrt{2}}$, rounded to the nearest half, is $3.5$. $a$ and $b$ are the integers on either side of $3.5$.
Now repeat the process with $n = c^2$, to be precise, $n = 25$:
$\sqrt{25(1 + \sqrt{2})^4}$ rounded down, squared, makes $841$, so $c = 29$.  $\dfrac{c}{\sqrt{2}}$ rounded to the nearest half is $20.5$, so $a = 20, b = 21$
Again, repeat with $n = 841$ and keep repeating as long as you like, and you will find all of the infinite Pythagorean triples where $a^2 + b^2 = c^2$ and $b = a + 1$
Here's a really big one I found:
$a = 666554398276487455279750313289061631639$
$b = 666554398276487455279750313289061631640$
$c = 942650270102046130733437746596776286089$
I don't know if this is the kind of answer you were looking for, but I thought I'd share my method anyway.
Regards,
Josiah Hamilton
Edit: I had written "rounded to the nearest integer" but it was supposed to say "rounded DOWN ..."
