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Is there a compendium of well-known (and elementary) Fibonacci/Lucas Number congruences? I've proven the following and would like to know if it is (a) trivial, (b) well-known, or (c) possibly new. $$ L_{2n+1} \pm 5(F_{n+2} + 1) \equiv (-1)^{n} \mod 10. $$ Is there a simple proof of this identity, requiring only basic identities of the two sequences? Any help or hints to this effect are certainly appreciated!

More generally, for integers $l,m,n$, we have $$ L_{12l + m+n} \pm 5(F_{60l + m} + F_{60l+ n} + F_{60l+ \text{gcd}(m,n)}) \equiv (-1)^{n} L_{12l + m - n} \mod 10. $$

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I have an easy proof of your congruence. I use only the following relations $L_{n+1}=L_n+L_{n-1},\ L_{2n}=L_n^2-2(-1)^n, L_n^2=5F_n^2+4(-1)^n$. Proceed as follows: $L_{2n+1}=L_{2(n+1)}-L_{2n}=L_{n+1}^2-L_n^2+4(-1)^n=$ $=5(F_{n+1}^2-F_n^2)+4(-1)^{n+1}-4(-1)^n+4(-1)^n=5F_{n+2}(F_{n+1}-F_n)-4(-1)^n$.

Therefore $L_{2n+1}\pm 5(F_{n+2}+1)\equiv (-1)^n (mod\ 10)$ is equivalent to $5F_{n+2}(F_{n+1}-F_n\pm 1)\equiv 5(-1)^n \mp 5 \equiv 0(mod\ 10)$.

We just have to prove that $(F_{n+1}-F_n\pm 1)F_{n+2}$ is even, which is clear, because $F_{n+2}\equiv F_{n+1}-F_n (mod\ 2)$.

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You might want to take a look at Thomas Koshy's Fibonacci Numbers with Applications, if you have access to a library that has it.

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  • $\begingroup$ Thank you. I just looked on Amazon in both chapters involving congruences and these do not appear in either of them. $\endgroup$ – user02138 May 24 '11 at 19:24
  • $\begingroup$ The identities that I see that are somewhat related are the following: $L_{12 l + n} \equiv L_{n} \mod 10$ and $F_{60 l + n} \equiv F_{n} \mod 10$ for $l,n \geq 0$. $\endgroup$ – user02138 May 24 '11 at 19:28

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