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

Question

I was searching for pythagorean triples where $b=a+1$, and I found using a python program I made the first 10 integer solutions:

1. $0^2+1^2=1^2$
2. $3^2+4^2=5^2$
3. $20^2+21^2=29^2$
4. $119^2+120^2=169^2$
5. $696^2+697^2=985^2$
6. $4059^2+4060^2=5741^2$
7. $23660^2+23661^2=33461^2$
8. $137903^2+137904^2=195025^2$
9. $803760^2+803761^2=1136689^2$
10. $4684659^2+4684660^2=6625109^2$

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?

Answer 1

I. Silver ratio

What you have discovered is the square of the silver ratio,

$$S=1+\sqrt{2} = 2.414213\dots$$

It is a cousin of the golden ratio, $\phi=\frac{1+\sqrt{5}}{2}$, and share similar properties. Your Pythagorean triple can be expressed as,

$$\Big(\frac{b-1}{2}\Big)^2+\Big(\frac{b+1}{2}\Big)^2=c^2\tag1$$

and factoring (WA link), we get the condition,

$$b^2-2c^2=-1$$

This is a Pell equation, and given one integer solution we can find an infinite more (this link, eqn. 35,36). Since this is the negative Pell equation, we use odd powers $n$,

$$\begin{aligned} b_n &= \frac{S^n+(-S)^{-n}}{2} = 1,\, 7,\, 41,\, 239,\dots\\ c_n &= \frac{S^n-(-S)^{-n}}{2\sqrt{2}} = 1,\, 5,\, 29,\, 169,\dots \end{aligned}\tag2$$

The $b_n$ is always odd, so $(1)$ are integers.

II. Answer:

Note that the contribution of $(-S)^{-n} \approx (-0.4142)^n$ in $(2)$ rapidly diminishes as $n$ becomes large. Thus,

$$c_n\approx \frac{S^n}{2\sqrt{2}}$$

Taking the ratio of $c_n$ for successive odd powers $n$,

$$ \frac{c_{2m+3}}{c_{2m+1}} = \Big( \frac{S^{2m+3}}{2\sqrt{2}} \Big)\Big(\frac{2\sqrt{2}}{S^{2m+1}}\Big)= S^2=(1+\sqrt{2})^2 = 5.828427\dots \tag3$$

and we recover your observation.

III. Neat stuff

Just like the golden ratio can be found in the pentagon, the silver ratio $S$ is in the octagon,

$\hskip2.8in$

It is also an infinitely nested radical,

$$S = 1+\sqrt{2} = 2\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{2^2}+\sqrt{\frac{1}{2^2}+\dots}}}$$

and in that nice continued fraction above using $e^{-\pi}$, as well as in a pi formula, etc, etc.