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


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 $F_{n+2}\equiv F_{n+1}-F_n (mod\ 2)$.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.