Only valid for Pythagoraean triples $\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$? $$\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2+\frac{b}{\sqrt2\cdots}}}=\sqrt{c+a}$$
Where (a,b,c) are the Pythagoraean Triples and are satisfy by the Pythagoras theorem $a^2+b^2=c^2$
An example of Pythagoraean triple (3,4,5)
It is true that this continued fraction is only valid for Pythagoraean Triple only?
I try other numbers and it seem that the only numbers that are valid are the Pythagoraean Triples. Can anyone verify this? Or show some examples where is also work for other numbers.

Let $\displaystyle x=\sqrt{2}+\frac{b}{\sqrt{2}+\ldots}$, then
\begin{align*}
  \sqrt{2}+\frac{b}{x} &= x \\
  x\sqrt{2}+b &= x^{2} \\
  x^{2}-\sqrt{2} \, x-b &= 0 \\
  x &= \frac{\sqrt{2}+\sqrt{2+4b}}{2} \\
  &= \sqrt{ \left( \frac{\sqrt{2}+\sqrt{2+4b}}{2} \right)^{2} } \\
  &= \sqrt{b+1+\sqrt{2b+1}} \\
\end{align*}
Take $a=\sqrt{2b+1}$ and $c=b+1$, it follows that $a^{2}+b^{2}=c^{2}$.
Note that the choices of $a$ and $c$ still have one degree of freedom if $(a,b,c)$ is not restricted to be a Pythagorean triple.
