Best Rational Approximations to Root 2 generate Pythagorean Triples/Proof 2

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

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

\(\ds \)

\(=\)

\(\ds \paren {P_{n + 1}^2 - P_n^2}^2 + \paren {2 P_{n + 1} P_n}^2\)

$\blacksquare$