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I'm trying to prove that$F_{5n+3}\text{mod}10 = L_{n}\text{mod}10$.

I rearranged it into a more solvable form of $F_{5n+3}-L_n = 10k$ (because if two numbers end in the same digit, their difference must be 10 and vice versa).

The problem is when I try to convert the Lucas numbers to Fibonacci numbers using almost any method, I get $F_{5n+3} - F_{n+1} - F_{n-1} = 10k$ or something similar. I need to find some way to get the coefficients to be equal so I can recombine them into simpler terms.

I'd appreciate any advice.

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Hint: Both Lucas numbers and Fibonacci numbers have periodic residues modulo any number. Now just find the periods and residues for both...

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F_n mod 10 has a period of 60, and L_n mod 10 has a period of 12, but how does that help me? – JShoe Jan 16 '14 at 15:25
Then F_{5n+3} has period twelve as L_{n} does and it is sufficient to check for first twelve $n$. As you don't need to compute numbers themselves, only residues this can be done. – user68061 Jan 16 '14 at 15:29
I've actually already done that, I'm just trying to follow up with an algebraic proof. – JShoe Jan 16 '14 at 15:36

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