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We have defined (in class) the Lucas number by $L_n = F_{n-1} + F_{n+1}$, where $F_i$ is the $i + 1$th Fibonacci number ($F_0 = 0, F_1 = 1$).

I have to prove that if $2 \mid n$ and $3\not\mid n$, then $L_n$ has a prime factor congruent to $3 \mod 4$. I have absolutely no idea on how to start this: I first thought about using induction, but I have no idea in this case how to prove my induction step, since I don't see how to write $L_n$ in function of some $L_k$'s on which I can use my induction hypothesis.

A second thought was that I could prove the contraposition, so assuming all possible prime factors are congruent with $2 \mod 4$ or $1 \mod 4$. However, in this case I am stuck on how to link this to the number $n$ and its divisors.

Any hints would be appreciated.

Student
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2 Answers2

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It is enough to prove that $L_{6n+2}$ and $L_{6n+4}$ are numbers of the form $4k-1$: in such a case they are odd numbers, hence they only have odd prime factors, and it is not possible that every prime factor is $\equiv 1\pmod{4}$, because otherwise $L_{6n+2}$ (or $L_{6n+4}$) would be $\equiv 1\pmod{4}$ too.

The Lucas numbers $\!\!\pmod{4}$ exhibit the following pattern: $$\begin{array}{|c||c|c|c|c|c|c|c|c|}\hline n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline L_n & 2 & 1 & 3 & 4 & 7 & 11 & 18 & 29 \\ \hline L_n\pmod{4} & 2 & 1 & \color{red}{3} & 0 & \color{red}{3} & 3 & 2 & 1 \\ \hline\end{array} $$ with period $6$. Since $L_2\equiv L_4\equiv 3\pmod{4}$, the claim follows.

Jack D'Aurizio
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  • First of all: thank you for your answer. If i get this right, I should first prove that the numbers $L_{6n +2}$ and $L_{6n+4}$ are of the form $4k-1$ and then I am finished? – Student Mar 21 '17 at 19:48
  • @Student: you are correct on the whole line. $2\mid n$ and $3\nmid n$ imply that $n$ is of the form $6k+2$ or $6k+4$: that might be obvious or not according to your knowledge, but in any case it is simple to prove. – Jack D'Aurizio Mar 21 '17 at 19:54
  • Once you prove that both $L_{6n+2}$ and $L_{6n+4}$ are of the form $4k+3$ you are finished. – Jack D'Aurizio Mar 21 '17 at 19:54
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    Yes, I realised that it is not that hard indeed (about $2\mid n$ and $3\not\mid n$). Thank you very much, I'll try to write this down and will accept your answer afterwards. – Student Mar 21 '17 at 19:56
  • I have encountered the same problem as with the hint in the other answer: I get stuck in trying to prove that the Lucas numbers exhibit the given pattern $\mod 4$: if i try to write the term $L_{6n + 2}$ in terms of $L_{k}$ with $k \equiv 2 \mod 6$, I get other terms of which I know nothing (in terms of the induction hypothesis). Are you aware of some relation between the $L_i$ which (i do not seem to be able to proof) is useful in proving the pattern? – Student Mar 21 '17 at 20:07
  • @Student: Lucas numbers fulfill the same recurrence relation as Fibonacci numbers, $L_{n+2}=L_{n+1}+L_{n}$, and such recurrence relation stays the same $\pmod{4}$, too. So, once you know that $L_0=2$ and $L_1=1$, you know the whole Lucas numbers sequence $\pmod{4}$, and it is simple to check that the period is $6$ by computing $L_n\pmod{4}$ for $n\in[0,8]$. Since $L_6\equiv L_0\pmod{4}$ and $L_7\equiv L_1\pmod{4}$, the pattern $2,1,3,0,3,3$ repeats forever. – Jack D'Aurizio Mar 21 '17 at 20:13
  • I am really sorry to keep bothering you about this: I was not aware that this holds for any recurrence relation, but if I would prove that the Lucas numbers are cyclic modulo some number $n$ and then show that for $n = 4$ I get some cycle, would this suffice in your opinion? – Student Mar 21 '17 at 20:34
  • @Student: of course it would. For an inspiration, you may also have a look at this famous answer of mine: http://math.stackexchange.com/a/872077/44121 – Jack D'Aurizio Mar 21 '17 at 20:37
  • Thank you very much, that answer seems to be very famous indeed :) – Student Mar 21 '17 at 20:40
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Since you're asked about the factorizations mod 4, start by looking at the Lucas numbers themselves mod 4. You get the repeating sequence

$1, 3, 0, 3, 3, 2, 1, 3, 0, 3, 3, 2, \ldots$

Can you prove this pattern holds? What does this imply about $L_n$ when $n$ is divisible by 2 but not by 3?

Michael Lugo
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  • Thank you! I will try this out and update my question with the solution I find using your hint. – Student Mar 21 '17 at 16:12
  • I have tried to prove this patern by induction but I am struggelling in writing $L_{6n}$ in terms of lucas numbers which differ 6 places. Anyways, thanks for the hint – Student Mar 21 '17 at 16:36