Best Rational Approximations to Root 2 generate Pythagorean Triples
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 odd term of $\sequence S$ can be expressed as:
- $\dfrac {2 a + 1} b$
such that:
- $a^2 + \paren {a + 1}^2 = b^2$
Proof 1
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.
$\blacksquare$
Proof 2
From Pell Number as Sum of Squares, we have:
- $P_{2 n + 1} = P_{n + 1}^2 + P_n^2$
Therefore:
\(\ds P_{2 n + 1}^2\) | \(=\) | \(\ds \paren {P_{n + 1}^2 + P_n^2}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds P_{n + 1}^4 + 2 P_{n + 1}^2 P_n^2 + P_n^4\) | Square of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds P_{n + 1}^4 + 2 P_{n + 1}^2 P_n^2 + P_n^4 + 2 P_{n + 1}^2 P_n^2 - 2 P_{n + 1}^2 P_n^2\) | adding zero | |||||||||||
\(\ds \) | \(=\) | \(\ds P_{n + 1}^4 - 2 P_{n + 1}^2 P_n^2 + P_n^4 + 4 P_{n + 1}^2 P_n^2\) | regrouping | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {P_{n + 1}^2 - P_n^2}^2 + \paren {2 P_{n + 1} P_n}^2\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$