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

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

    {2}^{p}+3^{p}={a}^{n} where p,a and n are positive integers \Longrightarrow 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?
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    Re: Guess: 2^p+3^p=a^n, then n=1 for ANY positive integer p. Can it be proved?

    well exactly as in the earlier problem, we know that p = 5,15 mod 20 (if it exists).
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    Re: Guess: 2^p+3^p=a^n, then n=1 for ANY positive integer p. Can it be proved?

    Quote Originally Posted by Deveno View Post
    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: 2*3*{5}^{4}*89*283*8971. Hence it's easy to see n=1, although p=5 (mod 25).
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    Re: Guess: 2^p+3^p=a^n, then n=1 for ANY positive integer p. Can it be proved?

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

    Quote Originally Posted by Deveno View Post
    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.
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    Re: Guess: 2^p+3^p=a^n, then n=1 for ANY positive integer p. Can it be proved?

    oh my goodness, how embarassing! you are quite correct. the even case would have to be handled separately.
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    Re: Guess: 2^p+3^p=a^n, then n=1 for ANY positive integer p. Can it be proved?

    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
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