<s>[INST] Odd convergents of $\sqrt 2$ and Pythagorean triangles with consecutive legs I would like to prove the following:

Take any odd convergent of $\sqrt 2$. The denominator gives the hypotenuse of a triangle; the numerator split into two consecutive integers gives the other two sides.

For example, the first $10$ convergents of $\sqrt 2$ are:
$$1,\quad \frac 3 2,\quad \frac 7 5,\quad \frac {17} {12},\quad \frac {41} {29},\quad \frac {99} {70},\quad \frac {239} {169},\quad \frac {577} {408},\quad \frac {1393} {985},\quad \frac {3363} {2378}$$
By looking at odd convergents, we see that


*

*$7 = 3 + 4$ and $5^2 = 3^2 + 4^2$;

*$41 = 20 + 21$ and $29^2 = 20^2 + 21^2$;

*$239 = 119 + 120$ and $169^2 = 119^2 + 120^2$;

*and so on.


Now, the convergents are given by $h_n / k_n$ with
$$h_1 = 1 \qquad h_2 = 3 \qquad h_{n+2} = 2 h_{n+1} + h_n$$
$$k_1 = 1 \qquad k_2 = 2 \qquad k_{n+2} = 2 k_{n+1} + k_n$$
and I have noticed that the statement is equivalent to the fact that
$$h_{2n+1}^2 - 2 k_{2n+1}^2 = -1$$
One way to prove it is by solving both recurrence relations, substituting the closed formulas and checking that the equality above holds for any $n$. This is quite long and involves some tedious computations.
Another way is by induction, but I believe one needs to strengthen the inductive hypothesis, since the equality involves only odd indices whereas the recurrence relation of $h_n, k_n$ involves both odd and even indices. I'm not sure how to do that.
Are there any alternative ways to prove the statement? A geometric proof would be particularly interesting, if there is one.
 [/INST] Consecutive Legs
In this answer it is shown that

Theorem: Let $m$ and $n$ be positive integers so that
  $$
\begin{align} 
&m\gt n\\
&m+n\text{ is odd}\\
&m\text{ and }n\text{ are relatively prime}
\end{align}
$$
  Then,
  $$
\begin{align}
a &= m^2 - n^2\\
b &= 2mn\\
c &= m^2 + n^2
\end{align}
$$
  gives all positive, relatively prime $a$, $b$, and $c$ so that
  $$
a^2 + b^2 = c^2
$$

To get consecutive legs, we need
$$
m^2-2mn-n^2=\pm1\tag1
$$
which means
$$
\left(\frac mn\right)^2-2\frac mn-1=\pm\frac1{n^2}\tag2
$$
and consequently, for $\frac mn\gt3-\sqrt2$,
$$
\begin{align}
\left|\,\frac mn-1-\sqrt2\,\right|
&=\frac1{n^2}\frac1{\frac mn-1+\sqrt2}\\
&\lt\frac1{2n^2}\tag3
\end{align}
$$
which requires that $\frac mn$ be a convergent of the continued fraction for $1+\sqrt2=(2;2,2,2,2,\dots)$:
$$
\left\{\vphantom{\frac21}\right.\underset{\begin{array}{c}\downarrow\\(3,4,5)\end{array}}{\frac21},\underset{\begin{array}{c}\downarrow\\(21,20,29)\end{array}}{\frac52},\underset{\begin{array}{c}\downarrow\\(119,120,169)\end{array}}{\frac{12}5},\underset{\begin{array}{c}\downarrow\\(697,696,985)\end{array}}{\frac{29}{12}},\underset{\begin{array}{c}\downarrow\\(4059,4060,5741)\end{array}}{\frac{70}{29}},\underset{\begin{array}{c}\downarrow\\(23661,23660,33461)\end{array}}{\frac{169}{70}},\quad\dots\left.\vphantom{\frac21}\right\}\tag4
$$

Computing the Pythagorean Triples
Solving the linear recurrence satisfied by numerators and denominators, $x_n=2x_{n-1}+x_{n-2}$, we get
$$
\begin{align}
m_k&=\frac{\left(1+\sqrt2\right)^{k+1}-\left(1-\sqrt2\right)^{k+1}}{2\sqrt2}\\
n_k&=\frac{\left(1+\sqrt2\right)^k-\left(1-\sqrt2\right)^k}{2\sqrt2}
\end{align}\tag5
$$
Therefore, we get the Pythagorean Triples
$$
\begin{align}
a_k&=\frac{\left(1+\sqrt2\right)^{2k+1}+2(-1)^k+\left(1-\sqrt2\right)^{2k+1}}4\\
b_k&=\frac{\left(1+\sqrt2\right)^{2k+1}-2(-1)^k+\left(1-\sqrt2\right)^{2k+1}}4\\
c_k&=\frac{\left(1+\sqrt2\right)^{2k+1}-\left(1-\sqrt2\right)^{2k+1}}{2\sqrt2}
\end{align}\tag6
$$
Note that
$$
\begin{align}
a_k-b_k&=(-1)^k\\
a_k+b_k&=m_{2k+1}-n_{2k+1}\\
c_k&=n_{2k+1}
\end{align}\tag7
$$
$(7)$ makes sense since $\frac{m-n}n$ is a convergent for $\sqrt2$ when $\frac{m}n$ is a convergent for $1+\sqrt2$. This proves the relations mentioned in the question.
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