TITLE: Pell Numbers and the Primes
QUESTION [6 upvotes]: I have observed that the Pell Numbers: a(0) = 1, a(1) = 2; for n > 1, a(n) = 2*a(n-1) + a(n-2)
and the related integer sequence:  a(0) = 2, a(1) = 0; for n > 1, a(n) = 2*a(n-1) + a(n-2)
together share an interesting property.
For any index number 'n', if n is a prime number, then it divides exactly into a(n) in just one, and only one, of the two sequences.  If n does not divide into a(n) in either sequence, then n is non-prime.
The sequences, and the primes that they generate, begin thus:
1  2  5  12  29  70…    3  5  11  13  19  29 …
2  0  2  4   10  24 …   2  7  17  23  31  41 …
I would like to find further information on this, such as who first discovered it, and whether it has been proved.
The sequence 2, 0, 2, 4, 10, 24, ...  is not listed in OEIS, so this may not be very well known.

REPLY [3 votes]: Your claim is true. Note that we can solve the recurrences to give
$$a_p=\frac{1}{\sqrt{2}}\left((1+\sqrt{2})^{p+1}-(1-\sqrt{2})^{p+1}\right)$$
for the first sequence. So for a prime $p$, we have that $p$ divides $a_p$ iff $p$ divides $2+2^{\frac{p+1}{2}}$, that is iff $2$ is a quadratic nonresidue modulo $p$, or in other words $p$ is $\pm 3 \pmod{8}$. (I used the fact that $\binom{p+1}{k}\equiv 0\pmod{p}$ for $2\le k\le p-1$.)
Similar analysis shows that for the second sequence we have $$a_p=\sqrt{2}\left((1+\sqrt{2})^{p-1}-(1-\sqrt{2})^{p-1}\right),$$
and so $a_p\equiv 2^{\frac{p+1}{2}}-2\pmod{p}$. So in the second sequence $p$ divides the $p$th term iff $2$ is a quadratic residue mod $p$, or in other words $p$ is $\pm 1\pmod{8}$.