Timeline for Which Fibonacci numbers are the sum of two squares?
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
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| Oct 25, 2012 at 15:14 | comment | added | joro | Are there congruence obstructions F_{10n} or F_{20n} to be sum of 2 squares? | |
| Oct 25, 2012 at 14:56 | answer | added | joro | timeline score: 2 | |
| Jun 23, 2011 at 20:44 | answer | added | Kristal Cantwell | timeline score: 0 | |
| Jun 23, 2011 at 19:23 | history | edited | Kevin O'Bryant | CC BY-SA 3.0 |
fixed tex issues
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| Jun 23, 2011 at 18:50 | history | edited | Kevin O'Bryant | CC BY-SA 3.0 |
added more data
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| Jun 17, 2011 at 12:20 | history | edited | Kevin O'Bryant | CC BY-SA 3.0 |
added 242 characters in body
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| Jun 16, 2011 at 21:35 | history | edited | Kevin O'Bryant | CC BY-SA 3.0 |
noted some basic facts
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| Jun 16, 2011 at 20:42 | history | edited | Kevin O'Bryant | CC BY-SA 3.0 |
Added examples
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| Jun 13, 2011 at 22:13 | comment | added | Mark Bennet | I wanted to double-vote this question, and I find the discussion of it really interesting. | |
| Jun 13, 2011 at 19:07 | answer | added | Will Jagy | timeline score: 5 | |
| Jun 13, 2011 at 18:03 | answer | added | J.C. Ottem | timeline score: 12 | |
| Jun 13, 2011 at 17:52 | comment | added | Eric Naslund | @Junkie: This is true, and not hard to prove. See my updated answer. It follows directly from the fact that if a prime of the form $5+6k$ divides $n$, then a prime of the form $3+4k$ divides $F_{2n}$, combined with the fact that the integers not divisible by primes of the form $5+6k$ have density $0$. | |
| Jun 13, 2011 at 5:46 | comment | added | Junkie | I think it was an Olympiad problem once: show that the set of $n$ such that $F_n$ has a prime divisor that is 3 mod 4 has (natural) density 1/2. None of the odd $n$ have such a divisor from your square relation of above, and then one can show that almost all (but not all) the even $n$ have such a divisor. | |
| Jun 13, 2011 at 2:05 | answer | added | Will Jagy | timeline score: 5 | |
| Jun 13, 2011 at 0:32 | comment | added | Vladimir Dotsenko | L_{30} is kinda irrelevant for our purposes though, since we are only interested in $2n\equiv 2\pmod{4}$ (because of divisibility by $3=F_4$ issues). | |
| Jun 12, 2011 at 23:01 | comment | added | Will Jagy | Also $L_{30}$ is the sum of two squares, go figure. | |
| Jun 12, 2011 at 22:58 | comment | added | Will Jagy | Note, the above is after Dror's comment on $F_{2n}$ and $F_n,$ as $F_{2n} = F_n L_n.$ Note that $L_{25}$ has a factor of 11, therefore $F_{50}.$ | |
| Jun 12, 2011 at 22:54 | comment | added | Will Jagy | From Eric Naslund's erased comments, it also means that Lucas number $L_n$ is the sum of two squares. For $n \leq 91,$ this is true for $n \equiv 1 \pmod 6$ itself prime or the product of such primes. However, $L_{97} = 3299 \cdot 56678557502141579.$ | |
| Jun 12, 2011 at 22:21 | comment | added | Dror Speiser | Small observation: since $F_{2n}=(2F_{n-1}+F_n)F_n$, and $\text{gcd}(2F_{n-1}+F_n,F_n)|2$, this implies that if $F_{2n}$ is a sum of squares then so is $F_n$, as well as $2F_{n-1}+F_n=F_{n-1}+F_{n+1}$. | |
| Jun 12, 2011 at 21:24 | answer | added | Eric Naslund | timeline score: 14 | |
| Jun 12, 2011 at 21:20 | comment | added | Kevin Buzzard | This question might be hard. If you really want to know the answer you could try a probabilistic model (which would prove nothing but might give you a good guess). I think probably $n=2$ mod 12 is the only case you care about (as long as $n>12$) and then you get lucky with 14,26,38 but are unlucky with 50 because there's a factor of 11. You want to try and guess the probability that a large fib number with $n=2 mod 12$ is divisible by an odd power of 11, 19, ... and then take the product and hope :-) | |
| Jun 12, 2011 at 20:55 | history | asked | Kevin O'Bryant | CC BY-SA 3.0 |