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Are there solutions to 4m^(n)=n^(2m) with m,n in Z+

  1. Apr 14, 2015 #1
    Hi, thanks for taking the time to read.
    To gain insight into [tex]n^m+n^m+1[/tex] and when it's prime, I looked at one case where it would be composite.
    I equated it to [tex](a+1)^2[/tex] and then substituted out to get [tex]4m^n=n^{2m}[/tex] Using mathematica, I was unable to get a solution. So here's my question: are there any solutions to it with integer n and m?
  2. jcsd
  3. Apr 14, 2015 #2


    Staff: Mentor

    By inspection, n=1 and m=2 or is that 2m an exponent?
  4. Apr 14, 2015 #3
    2m is an exponent. I fixed it.
  5. Apr 14, 2015 #4


    Staff: Mentor

    There's likely no integer solution. In the simpler case of x^y = y^x the only integer solution is 2 and 4.

    My pocket CAS on iOS couldnt find any solutions either but it couldnt solve my easier one either.
  6. Apr 14, 2015 #5
    Well that's pretty interesting since no integer solutions would imply that [tex]x^y+y^x+1[/tex] will never be a perfect square. Thank you.

    Edit: Found m=1 and n=2 is a solution. I wonder if that's the only integer solution.
    Last edited: Apr 14, 2015
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