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Fibs and Mods. (Posted on 2016-03-30) Difficulty: 2 of 5

Consider the Fibonacci numbers , F(n) mod N. At some point, N divides F(n) for the first time. Eventually, at F(n+m), N divides F(n+m) for the second time.

Claim 1: F(n+m) /F(n) is always a Lucas number, L(k).

Examples:
N=2: F(n) = 2, F(n+m) = 8; 8/2 = 4 = L(3)
N=3: F(n) = 3, F(n+m) = 21; 21/3 = 7 = L(4)
N=4: F(n) = 8, F(n+m) = 144: 144/8 = 18 = L(6)
N=5: F(n) = 5, F(n+m) = 55: 55/5 = 11 = L(5)
N=6: F(n) = 144, F(n+m) = 46368: 46368/144 = 322 = L(12)
N=7: F(n) = 21, F(n+m) = 987: 987/21 = 47 = L(8)
etc.

It looks as though those Lucas numbers are hopping about, so let's instil some discipline to the order of their appearance:

Claim 2: F(n+m) /F(n) =L(m).

Prove both claims, or find counter-examples.

See The Solution Submitted by broll    
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  Subject Author Date
Hints/Tipsno proof here, just some verification, and a new factCharlie2016-03-30 11:28:53
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