Lucas-number prime factor form proofs

[Raman]

Originally Posted by Batalov

Lucas1249 c262 = p115 * p147

p115 = 1230907246701748850213178915950086177557919307463961418238191238338563780421891858424331033928188919064775254538919
p147 = 861662799748056902967441789531902541845512917066647559276041990818216830987090087949122773698681932527206937859817129749788517846169149201190846679

B+D

How to Prove that any prime factor for L_p ≡ 1 (mod p)
How to Prove that any prime factor for L_p ≡ 1, 9 (mod 10)

[Raman]

Originally Posted by Raman

How to Prove that any prime factor for L_p ≡ 1 (mod p)
How to Prove that any prime factor for L_p ≡ 1, 9 (mod 10)

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



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?,

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

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, why?,
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, why?,
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, why?,