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