mersenneforum.org - Lucas-number prime factor form proofs

[QUOTE=Raman;301695]How to Prove that any prime factor for L[SUB]p[/SUB] ≡ 1 (mod p)
How to Prove that any prime factor for L[SUB]p[/SUB] ≡ 1, 9 (mod 10)[/QUOTE]
Thus, it does so thereby, similar thing holds always for the Fibonacci numbers, candidates again with these Following examples

For this example, consider with the following statements, in the fact, in the turning process, all at once

[tex]F_p = \begin{cases} 0\ (mod\ p) & \mbox{if } p = 5 \\ 1\ (mod\ p) & \mbox{if } p\ \equiv\ 1,\ 4\ (mod\ 5) \\ -1\ (mod\ p) & \mbox{if } p\ \equiv\ 2,\ 3\ (mod\ 5). \\ \end{cases} [/tex]

any prime factor for F[sub]p[/sup] ≡ ±1 (mod p), p ≠ 5.

On the other hand,

any prime factor for F[sub]p[/sup] ≡ 1 (mod 4), why?

i.e. all the values for F[sub]n[/sup] for all the odd values for the literal n,

are being the sum of two squares, why?[STRIKE],[/STRIKE]

any prime factor for 2[sub]n[/sup]-1, for odd n ≡ 1, 7 (mod 8)[STRIKE], why?[/STRIKE]
any prime factor for 2[sub]n[/sup]+1, for odd n ≡ 1, 3 (mod 8)[STRIKE], why?[/STRIKE]
any prime factor for 2[sub]n[/sup]+1, for even n ≡ 1, 5 (mod 8)[STRIKE], why?[/STRIKE]

i.e. all the values for 2[sub]n[/sup]-1 for all the odd values for the literal n, aren't being the sum of two squares, [STRIKE]why?,[/STRIKE]
i.e. all the values for 2[sub]n[/sup]+1 for all the odd values for the literal n, aren't being the sum of two squares, [STRIKE]why?,[/STRIKE]
i.e. all the values for 2[sub]n[/sup]+1 for all the even values for the literal n, are being the sum of two squares, [STRIKE]why?,[/STRIKE]