Why do prime factors of odd term Lucas numbers only end in 1 or 9? I'm working on a problem, which I've eventually reduced to the following question: show that every odd term Lucas number has a prime factor that ends with either 1 or 9. Here the Lucas sequence is defined by the recursion: $$a_0=2, a_1=1, a_{n+1}=a_n+a_{n-1}.$$
I looked up a list of prime factorizations of Lucas numbers on the following page:
https://r-knott.surrey.ac.uk/Fibonacci/lucas200.html
Surprisingly, I find all the odd term Lucas numbers to have only prime factors that end with either 1 or 9, with the exception that $a_{6k+3}$ is also divisible by 4. It's easy to show using modulo arithmetic that all $a_{6k+3}$ are divisible by 4 and no higher powers of 2, while all other odd term Lucas numbers are odd. If we count out those factors of 4, all the other prime factors of the odd term Lucas numbers seem to end only with 1 or 9. This is a curious observation, but I don't know how one can prove this. Can anyone help please!

I made another Conway Topograph diagram for $x^2 - 5 y^2$  The $x,y$ pairs are written vertically in green.  The values of $x^2 - 5 y^2$ are in pink.
Right, see displayed $11^2 - 5 \cdot 5^2 = -4, \; \; $ $29^2 - 5 \cdot 13^2 = -4, \; \; $ $199^2 - 5 \cdot 89^2 = -4, \; \; $ $521^2 - 5 \cdot 233^2 = -4, \; \; $
Also, $38^2 - 5 \cdot 17^2 = -1, \; \; $  and doubling this gives
$76^2 - 5 \cdot 34^2 = -4, \; \; $

If $x$   is one of your alternate Lucas numbers, there is a Fibonacci number $y$ such that
$$  x^2 - 5 y^2 = -4  $$
From $x^2 = 5 y^2 - 4, $   we see $x^2 = 5 y^2 - 2^2 = - (2^2 - 5 y^2 )$    and
$$   x | 2^2 - 5 y^2   $$
Now if there is any prime $q$   with $q \equiv \pm 2 \pmod 5,$  and if $q | x,$    then
$$   q | 2^2 - 5 y^2   $$
However, for any such prime, Legendre symbol $(20|q) = -1.$  This forces $q | y $   along with $q | 2.$
Thus, it is allowed for $x$ to be even (divisible by 2)   and divisible by primes $p \equiv 1,4 \pmod 5,$   but not by $q \equiv \pm 2 \pmod 5$  when $q > 2$
On this site, you may have seen this fact:   for a prime $r \equiv 3 \pmod 4,$  if $r | u^2 + v^2,$   then both $r|u$ and $r|v$  This is a general thing about the Legendre symbol.
