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:

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?

Solution 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.

Solution 2:

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.)

Solution 3:

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.