Odd convergents of $\sqrt 2$ and Pythagorean triangles with consecutive legs

Question

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.

Answer 1

Assuming that $(a,a+1,c)$ is a Pythagorean triple we have that $2a^2+2a+1=c^2$ or $(2a+1)^2=2c^2-1$, so $(2a+1)$ belongs to the set of solutions of $x^2-2y^2=-1$ ($x=2a+1$). By the theory of Pell equations all the positive solutions are generated by the fundamental solution $(x_1,y_1)=(1,1)$ via $$ x_n+\sqrt{2} y_n = (1+\sqrt{2})(3+2\sqrt{2})^n, $$ so the sequence $\{x_n\}_{n\geq 1}$ is a linear recurrent sequence with characteristic polynomial $x^2-6x+1$ (the minimal polynomial of $3+2\sqrt{2}$ over $\mathbb{Q}$), i.e. $x_{n+2}=6x_{n+1}-x_n$. This is also the recurrence fulfilled by the sequence $\{p_{2n+1}\}_{n\geq 0}$, where $\frac{p_n}{q_n}$ is the $n$-th convergent of $\sqrt{2}$, due to $$ \sqrt{2}=[1;2,2,2,\ldots]. $$ A simple induction then proves the wanted claim.

Answer 2

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 $1+\sqrt2=(2;2,2,2,2,\dots)$: $$ \left\{\vphantom{\frac21}\right.\underset{\begin{array}{c}\downarrow\\(3,4,5)\end{array}}{\frac21},\underset{\begin{array}{c}\downarrow\\(21,20,29)\end{array}}{\frac52},\underset{\begin{array}{c}\downarrow\\(119,120,169)\end{array}}{\frac{12}5},\underset{\begin{array}{c}\downarrow\\(697,696,985)\end{array}}{\frac{29}{12}},\underset{\begin{array}{c}\downarrow\\(4059,4060,5741)\end{array}}{\frac{70}{29}},\underset{\begin{array}{c}\downarrow\\(23661,23660,33461)\end{array}}{\frac{169}{70}},\quad\dots\left.\vphantom{\frac21}\right\}\tag4 $$

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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.