Parity of Pell Numbers

Theorem
Consider the Pell numbers $P_0, P_1, P_2, \ldots$
 * $0, 1, 2, 5, 12, 29, \ldots$

$P_n$ has the same parity as $n$.

That is:
 * if $n$ is odd then $P_n$ is odd
 * if $n$ is even then $P_n$ is even.

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $P_n$ has the same parity as $n$.

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $P_0 = 0$

which is even.

$P \left({1}\right)$ is the case:
 * $P_1 = 1$

which is odd.

Thus $P \left({n}\right)$ holds for $n = 0$ and $n = 1$.

These comprise the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({j}\right)$ is true for all $j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\forall j \le k: P_j$ has the same parity as $j$

from which it is to be shown that:
 * $P_{k + 1}$ has the same parity as $k + 1$.

Induction Step
This is the induction step:

By definition of Pell numbers:


 * $P_{k + 1} = 2 P_k + P_{k - 1}$

$2 P_k$ is even by definition.

By Sum of Even Numbers is Even, if $P_{k - 1}$ is even then so is $P_{k + 1}$.

Similarly, if $P_{k - 1}$ is odd then so is $P_{k + 1}$.

Thus the parity of $P_{k + 1}$ matches the parity of $P_{k - 1}$.

But by the induction hypothesis, the parity of $P_{k - 1}$ matches $k - 1$.

Thus the parity of $P_{k + 1}$ matches $k + 1$.

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 0}: P_n$ has the same parity as $n$.