Pythagorean triplets of the form $a^2+(a+1)^2=c^2$ and the space between them 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?

Actually, you should write a general formula, and these issues are constantly emerging.
Solutions of the equation:  $$x^2+(x\pm{a})^2=z^2$$
Can be written using the solutions of Pell's equation: $p^2-2s^2=\pm{a}$
And have the form:
$$x=2s(p+s)$$
$$z=p^2+2ps+2s^2$$
If we introduce a replacement:
$$q=p^2+2ps+2s^2$$
$$t=2ps$$
Solutions have the form:
$$x=t^2-q^2$$
$$z=q^2+t^2$$
Solutions have the form:
$$x=2qt$$
$$z=q^2+t^2$$
It must be remembered that the number of $p,s$ can be of any sign.
