TITLE: Pythagorean triplets of the form $a^2+(a+1)^2=c^2$ and the space between them
QUESTION [22 upvotes]: I was searching for pythagorean triples where $b=a+1$, and I found using a python program I made the first 10 integer solutions:

$0^2+1^2=1^2$
$3^2+4^2=5^2$
$20^2+21^2=29^2$
$119^2+120^2=169^2$
$696^2+697^2=985^2$
$4059^2+4060^2=5741^2$
$23660^2+23661^2=33461^2$
$137903^2+137904^2=195025^2$
$803760^2+803761^2=1136689^2$
$4684659^2+4684660^2=6625109^2$

Now what's so interesting? I discovered that any $c$, divided by the previous (for example $5/1$ or $29/5$) limits to $5.828427...=\left(\frac{1}{\sqrt2-1}\right)^2=\sqrt8+3$. My question: why?

REPLY [2 votes]: This explains why the ratios converge to the square of the silver ratio.


This shows how all of the primitive Pythagorean triples can be generated by 
Pellian sequences, and therefore the ratios of consecutive hypotenuses from any of those sequences would converge to the square of the silver ratio. 
All of the primitive triples can be listed by sequences with the same recursion relation, just different initial values. The initial values will only affect the coefficients in the explicit forms for those sequences. This will not affect what their ratios converge to as n goes to infinity for each sequence.
THEOREM: All of the primitive Pythagorean triples can be generated, ordered and
largely sorted without redundancy by substitution into {v^2 – u^2, 2uv, v^2 + u^2 } of consecutive pairs of terms of the Pell numbers and similar sequences formed by the same recursion relation, P(n + 2) = 2P(n + 1) + P(n), and initial values n, n + m and n – m, 3n – 2m, where n and m are positive integers, n > m, gcd(m,n) = 1,
and m is odd. 
(The leg difference of triples generated from both sequences is d = 2n^2 – m^2. 
And the Pell numbers can be considered the special case of a singleton sequence: 
n, n + m, . . . , where n = m = 1.)