20150821, 03:18  #1 
Aug 2015
2×3^{3} Posts 
Fibonacci number as sum of cubes
Hello everyone!
Hopefully this is the right place to post. A while ago I asked on math.stackexchange if there was a Fibonacci number that was the sum of 2 positive cubes, besides 2. I am hoping you guys might have some new ideas. I am a beginner when it comes to number theory and math, but I am surprised to not be able to find any other research on this. 
20150821, 12:04  #2  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:
0,1,1,2,3,5,8,4,3,7,1,8,0,8,8,7,6,4,1,5,6,2,8,1,... so only positions that are 1,2,3,4,7,10,12,14,15,16,19,22,23,0 mod 24 can the sum of two cubes to begin with. and that's what I can come up with right now. Last fiddled with by science_man_88 on 20150821 at 12:09 

20150821, 12:22  #3  
Nov 2003
1110100100100_{2} Posts 
Quote:
Read H. Cohen's book on Diophantine Equations. Last fiddled with by R.D. Silverman on 20150821 at 12:30 

20150821, 12:31  #4 
Nov 2003
2^{2}·5·373 Posts 

20150821, 13:08  #5 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
not to mention you can use that fact that prime positions matter most because starting at 1,1 instead of 0,1 any position that isn't prime divides by a lower prime one and so the value the latter one divided by the previous one needs specific form for the sum of cubes to work ( a cube number after division is the easiest example).

20150821, 14:29  #6  
Nov 2003
16444_{8} Posts 
Quote:


20150821, 14:40  #7 
"Forget I exist"
Jul 2009
Dumbassville
20C0_{16} Posts 
not true for example 2 is at a prime position starting with 1,1 therefore with the alternate form of the Fibonacci sequence starting 1,1,2,3,5,8 ( note no 0) if any formula connecting the positions that are multiples of 3 together can be a cube infinitely often then there could be infinitely many of them contrary to what you said earlier. so in order to disprove infinitely many all you have to do is check the prime position ones assuming of course you can find a formula to link the fibonacci numbers that are in composite positions to the prime factors of that position then you have to disprove it can be a cube infinitely often. the only time you have to check positions that aren't prime would be if no formula that can be a cube infinitely often is found.

20150821, 15:12  #8  
Nov 2003
2^{2}×5×373 Posts 
Quote:


20150821, 15:56  #9  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:


20150821, 17:14  #10 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,109 Posts 
science_man_88, stop your
There is nothing that needs to be added to stackexchange's discussion which I am afraid that you were not bothered to read. 
20150821, 17:31  #11  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:


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