Thread: Guess: 2^p+3^p=a^n, then n=1 for ANY positive integer p. Can it be proved?

where p,a and n are positive integers n=1.

An earlier problem requires p to be prime numbers:
http://www.mathhelpforum.com/math-he...nt-183464.html

Here we drop the prime number requirement. Conclusion still true?

I'm just curious I know very few results in number theory! Do we already have ready results for this?

well exactly as in the earlier problem, we know that p = 5,15 mod 20 (if it exists).

Originally Posted by Deveno

well exactly as in the earlier problem, we know that p = 5,15 mod 20 (if it exists).

Well, maybe you missed the subtleties there, Deveno!

Consider this: let p=25, then 2^25+3^25=847322163875. It's prime factorization is: . Hence it's easy to see , although p=5 (mod 25).

i'm not saying every p of the form 5 (mod 20) or 15 (mod 20) will work, just that those are the only possible candidates.

in other words, we've narrowed down the numbers worth checking to 2 out of every 20.

Originally Posted by Deveno

i'm not saying every p of the form 5 (mod 20) or 15 (mod 20) will work, just that those are the only possible candidates.

in other words, we've narrowed down the numbers worth checking to 2 out of every 20.

2^p+3^p=0 (mod 5) iff p is odd number. So the whole proof there works only for p=odd numbers. Since p can be even numbers now, you can't even draw conclusions from there to say you need only check 2 out of 20.

oh my goodness, how embarassing! you are quite correct. the even case would have to be handled separately.

Hi, for those who are interested: I posted the question on Mathoverflow, check it out there ${2}^{p}+{3}^{p}={a}^{n}$ , then n=1 for any p ? - MathOverflow