Fibonacci number as sum of cubes

[jux]

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.

[science_man_88]

Originally Posted by jux

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.

cubes are 0 1 and 8 mod 9 sum any two and you can get 0 1 2 7 or 8 mod 9 every number has a pisano period , however lets look at what places these values mod 9 pop up. the Fibonacci sequence mod 9 repeats:

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.

[R.D. Silverman]

Originally Posted by jux

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.

There are at most finitely many.

Read H. Cohen's book on Diophantine Equations.

[R.D. Silverman]

Originally Posted by R.D. Silverman

There are at most finitely many.

Read H. Cohen's book on Diophantine Equations.

It is possible that an application of Baker's linear forms in logarithms might be able
to place an upper bound on the solutions. I am not an expert in this area.

[science_man_88]

Originally Posted by R.D. Silverman

There are at most finitely many.

Read H. Cohen's book on Diophantine Equations.

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).

[R.D. Silverman]

Originally Posted by science_man_88

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).

Complete word salad. Total nonsense.

[science_man_88]

Originally Posted by R.D. Silverman

Complete word salad. Total nonsense.

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.

[R.D. Silverman]

Originally Posted by science_man_88

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.

More gibberish.

[science_man_88]

Originally Posted by R.D. Silverman

More gibberish.

Originally Posted by https://en.wikipedia.org/wiki/Fibonacci_number#Divisibility_properties

Every 3rd number of the sequence is even and more generally, every kth number of the sequence is a multiple of Fk

therefore if there's a form for the composite positioned ones from the primitive ones then you only need to check anything that isn't prime only if there's not a possibility of the prime positioned ones leading to cubes infinitely often.

[Batalov]

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.

[science_man_88]

Originally Posted by Batalov

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.

I did read it