Best Rational Approximations to Root 2 generate Pythagorean Triples

From ProofWiki
Jump to navigation Jump to search

Theorem

Consider the Sequence of Best Rational Approximations to Square Root of 2:

$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$

Every other term of $\sequence S$ can be expressed as:

$\dfrac {2 a + 1} b$

such that:

$a^2 + \paren {a + 1}^2 = b^2$
$b$ is odd.


Proof

From Parity of Best Rational Approximations to Root 2‎:

The numerators of the terms of $\sequence S$ are all odd.
For all $n$, the parity of the denominator of term $S_n$ is the same as the parity of $n$.

Thus it follows that every other term of $\sequence S$ has a numerator and a denominator which are both odd.



This proof proceeds by induction.

Basis for the Induction

$\dfrac 1 1$ can be expressed as $\dfrac {2(0) + 1} 1$, and:

$0^2 + \paren {0 + 1}^2 = 1^2$


Induction Hypothesis

This is our induction hypothesis:

The best rational approximation $\dfrac {p_k} {q_k}$ of $\sqrt 2$, when expressed as:

$\dfrac {2 a + 1} b$

gives the relation:

$a^2 + \paren {a + 1}^2 = b^2$


We need to show that the best rational approximation $\dfrac {p_{k + 2} } {q_{k + 2} }$ of $\sqrt 2$, when expressed as:

$\dfrac {2 a' + 1} {b'}$

also give the relation:

$a'^2 + \paren {a' + 1}^2 = b'^2$


Induction Step

This is our induction step:

From the induction hypothesis we have:

$a = \dfrac {p_k - 1} 2, b = q_k$

and thus:

$\paren {\dfrac {p_k - 1} 2}^2 + \paren {\dfrac {p_k - 1} 2 + 1}^2 = q_k^2$

Expanding, we have:

$\dfrac {p_k^2} 2 + \dfrac 1 2 = q_k^2$


Now by Relation between Adjacent Best Rational Approximations to Root 2, we have:

$\dfrac {p_{k + 1} } {q_{k + 1} } = \dfrac {p_k + 2 q_k} {p_k + q_k}$

We check that, via GCD with Remainder:

$\gcd \set {p_k + 2 q_k, p_k + q_k} = \gcd \set {q_k, p_k + q_k} = \gcd \set {q_k, p_k} = 1$

Since both fractions are in canonical form and Canonical Form of Rational Number is Unique, we can write:

$p_{k + 1} = p_k + 2 q_k$
$q_{k + 1} = p_k + q_k$

Therefore:

$p_{k + 2} = 3 p_k + 4 q_k$
$q_{k + 2} = 2 p_k + 3 q_k$


We need to show that:

$\paren {\dfrac {p_{k + 2} - 1} 2}^2 + \paren {\dfrac {p_{k + 2} - 1} 2 + 1}^2 = q_k^2$

or:

$\dfrac {p_{k + 2}^2} 2 + \dfrac 1 2 = q_{k + 2}^2$


We have:

\(\ds \frac {p_{k + 2}^2} 2 + \frac 1 2\) \(=\) \(\ds \frac {\paren{3 p_k + 4 q_k}^2} 2 + \frac 1 2\)
\(\ds \) \(=\) \(\ds \frac {9 p_k^2 + 24 p_k q_k + 16 q_k^2} 2 + \frac 1 2\) Square of Sum
\(\ds \) \(=\) \(\ds \frac {p_k^2} 2 + \frac 1 2 + 4 p_k^2 + 12 p_k q_k + 8 q_k^2\)
\(\ds \) \(=\) \(\ds 4 p_k^2 + 12 p_k q_k + 9 q_k^2\)
\(\ds \) \(=\) \(\ds \paren {2 p_k + 3 q_k}^2\) Square of Sum
\(\ds \) \(=\) \(\ds q_{k + 2}^2\)

The result follows by induction.

$\blacksquare$


Sources