An Identity for Pell-numbers

Question

The Pell-numbers are defined recursively by:

$P_0 = 0, P_1 = 1$ and $ P_{n+2} = 2P_{n+1} + P_n$

I am stuck trying to prove the identity:

$P_{2n+1}^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$

A proof would be great, otherwise a list of known Pell-identies would also help

Answer 1

There is a well-know Pell-identity: $P_{2n+1} = P_n^2 + P_{n+1}^2$ which can be found in this paper. It can also be shown directly.

Using this, we just need to show: $(P_n^2 + P_{n+1}^2)^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$

This is done by induction: The statement clearly holds for $n = 0$, therefore suppose it holds for any $n$ and consider the following argument:

$(P_{n+1}^2 + P_{n+2}^2)^2 - 1 = (P_{n+1}^24P_{n+1}^2 + 4P_nP_{n+1} + P_n^2)^2 - 1$

$= 16P_{n+1}^2(P_n + P_{n+1})^2 + 8P_{n+1}(P_n + P_{n+1})(P_n^2 + P_{n+1}^2) + (P_n^2 + P_{n+1}^2)^2 -1$

$= 16P_{n+1}^2(P_n + P_{n+1})^2 + 8P_{n+1}(P_n + P_{n+1})(P_n^2 + P_{n+1}^2) + 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$

$= 4P_{n+1}(P_n + P_{n+1})(4 P_{n+1}(P_n + P_{n+1}) + 2 P_n^2 + 2 P_{n+1}^2 + P_n(P_{n+1} - P_n))$

$= 4P_{n+1}(P_n + P_{n+1})(P_n^2 + 5P_{n+1}P_n + 6P_{n+1})$

$= 4P_{n+1}(P_n + P_{n+1})(P_n + 2P_{n+1})(P_n + 3P_{n+1})$

At this point, we can just apply the fact that $P_n + P_{n+1} = P_{n+2}$, and $P_n + P_{n+1} = P_{n+2} - P_{n+1}$ and substitute that into the above equation:

$4P_{n+1}P_{n+2}(P_{n+2} - P_{n+1})((P_{n+2} + P_{n+1}) = 4P_{n+1}P_{n+2}(P_{n+2}^2 - P_{n+1}^2)$

which completes the induction.

Answer 2

You know $P_{n+2} = 2 P_{n+1} + P_{n}$ $\Rightarrow$ $P_{n+2}-(1+\sqrt{2})P_{n+1}= (1-\sqrt{2})(P_{n+1}-(1+\sqrt{2})P_{n})$. In other word, we have \begin{align*} \frac{P_{n+2}-(1+\sqrt{2})P_{n+1}}{P_{n+1}-(1+\sqrt{2})P_{n}} = (1-\sqrt{2}). \end{align*} Since $P_0 = 0, P_1 = 2, P_2 = 2$, we deduce that \begin{align*} P_{n} = \frac{(1+\sqrt{2})^n - (1-\sqrt{2})^n }{2\sqrt{2}}. \end{align*} You refer to Pell_number.