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Math Help - Solve in natural numbers

  1. #1
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    Solve in natural numbers

    Hi,

    I wonder whether somebody can advise me how to solve in natural numbers (or integers) the following equation:
    3^x+4^y=5^z
    I have spent a couple of weeks thinking about it without any considerable progress. The obvious guess is that the only solution is (x,y,z)=(2,2,2). I've looked at the equation modulo different numbers 3,4,5,9,16,25 ... etc, but so far that wasn't very conclusive in proving that (2,2,2) is the only solution. I've tried a lot of other different things with the equation with the same negative result. Honestly I've run out of ideas of how to solve the equation.

    Any help is appreciated.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    In my opinion, if the statement is true (I feel that it probably is), then it must be quite hard to prove.

    It would follow directly from Beal's conjecture but unfortunately it is not proven.
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  3. #3
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    Quote Originally Posted by segasai View Post
    Hi,

    I wonder whether somebody can advise me how to solve in natural numbers (or integers) the following equation:
    3^x+4^y=5^z
    I have spent a couple of weeks thinking about it without any considerable progress. The obvious guess is that the only solution is (x,y,z)=(2,2,2). I've looked at the equation modulo different numbers 3,4,5,9,16,25 ... etc, but so far that wasn't very conclusive in proving that (2,2,2) is the only solution. I've tried a lot of other different things with the equation with the same negative result. Honestly I've run out of ideas of how to solve the equation.

    Any help is appreciated.
    If you're solving over the integers (or your convention is that 0 is a natural number) then there's the additional solution (0,1,1).

    I ran a quick check up to z = 1000 and couldn't find any additional solutions. Good luck with the proof!
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  4. #4
    Senior Member Shanks's Avatar
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    Quote Originally Posted by pomp View Post
    If you're solving over the integers (or your convention is that 0 is a natural number) then there's the additional solution (0,1,1).

    I ran a quick check up to z = 1000 and couldn't find any additional solutions. Good luck with the proof!
    How do you do this?
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  5. #5
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    Quote Originally Posted by Shanks View Post
    How do you do this?
    I done it in Haskell: http://en.wikipedia.org/wiki/Haskell...mming_language)
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  6. #6
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    Dear segasai,

    I have attached a solution that I have achieved for this problem. Please go through it and reply me soon about your views regarding the solution. If you have any problem regarding the attachment please don't hesitate to reply me. I value your reply as I enjoyed solving the problem. I have corrected this attachments they are in the order page 2, page 3, page 1, page 4
    Hope this helps.
    Attached Thumbnails Attached Thumbnails Solve in natural numbers-2.jpg   Solve in natural numbers-3.jpg   Solve in natural numbers-1.jpg   Solve in natural numbers-4.jpg  
    Last edited by Sudharaka; January 5th 2010 at 01:52 AM.
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  7. #7
    Senior Member Shanks's Avatar
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    (0,0,0) is a solution? clearly faalse!
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  8. #8
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    Hi everyone,

    Sorry if I confused you. (0,0,0) is not a solution. Shame on me. And thank you very much Shanks for telling me that. But could anyone tell me whether my idea is correct or not neglecting this error. I have corrected the attachments.

    Thanks again Shanks.
    Last edited by Sudharaka; January 5th 2010 at 01:48 AM.
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  9. #9
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    Given that x,y,z \in \mathbf{N} we know that x>0.

    Then 3^x+4^y \equiv_3 1 and 5^z \equiv_3 (-1)^z. Therefore z=2a and we can write 3^x=5^{2a}-4^y=(5^a+2^y)(5^a-2^y).

    We see that both factors can't be a power of 3 since there sum isn't divisible by 3. Therefore we've the following cases:
    1) 5^a+2^y=3^x and 5^a-2^y=1
    2) 5^a-2^y=3^x and 5^a+2^y=1

    For the first case, using mod3 on the equality 5^a-2^y=1 implies that y must be even and a must be odd. This is true since 5^a-2^y \equiv_3 (-1)^a-(-1)^y\equiv_3 1.

    For y>2 we see that 5^a+2^y \equiv_8 5 (using that a is odd) but clearly 3^x \equiv_8 1 \mbox{ or } 3 so there is no solution for y>2

    We know that y is even so we need to consider y=0 and y=2.

    For y=0 we obtain 5^a+1=3^x which clearly has no solutions (use mod4).

    For y=2 we obtain 5^a-2^2=1 and this implies that a=1 which is z=2 and a solution is therefore (2,2,2)

    For the second case, 5^a+2^y=1 clearly can't be true so there are no new solutions.

    Therefore (2,2,2) is the only solution, as expected.


    If we add the possibilty for x,y,z to be 0 we've to consider x=0
    This gives us 1+4^y=5^z
    It's easy to see that y=z=1 is a solution.

    We have to consider y>1
    z must be odd since 1+4^y \equiv_3 2 and 5^z \equiv_3 2^z

    But using mod8 we see that 1+4^y \equiv_8 1 and 5^z \equiv_8 (-3)^z so clearly z must be even.

    We've reached a contradiction so there are no more solutions than (2,2,2) and (0,1,1) for x,y,z \in \{0,1,.. \}
    Last edited by DavidEriksson; January 5th 2010 at 04:12 AM. Reason: Adding the possibility for x,y,z to be 0
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  10. #10
    Senior Member Shanks's Avatar
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    I thought, Sudharaka's solution is correct for the case t'=11.
    Sudharaka, could you explain to us why you set t'=11, and can other case be reduced to the case t'=11(here I don't think so).
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  11. #11
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    Thank you, Sudharaka for the suggested solution.
    But as Shanks pointed, the assumption t'=11 is incorect. In fact I checked numerically, it is really incorrect. But I believe that your solution can be fixed ... even if t' is not const.
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  12. #12
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by DavidEriksson View Post
    For y>2 we see that 5^a+2^y \equiv_8 5 (using that a is odd) but clearly 3^x \equiv_8 1 \mbox{ or } 3 so there is no solution for y>2
    Why does this hold? I can't see how you get \equiv_8 5. I just get \equiv_8 5 + 4^{y'} where y' = \frac{y}{2}.

    Other than this and your omission of the x=0 case, it all looks good!

    This means I can stop running my maple program now. Pity. It has been on the go for over 6000 seconds now...
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  13. #13
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    Quote Originally Posted by Swlabr View Post
    Why does this hold? I can't see how you get \equiv_8 5. I just get \equiv_8 5 + 4^{y'} where y' = \frac{y}{2}.

    Other than this and your omission of the x=0 case, it all looks good!

    This means I can stop running my maple program now. Pity. It has been on the go for over 5000 seconds now...
    Hi!
    When using "natural numbers" 0 is usually excluded, but i can add the case to the solution.

    We know that a is odd and y is even. For y>2 it's easy to see that 8 \mid 2^y so 2^y \equiv_8 0 for y>2

    Since a is odd, a=2k+1 and thus 5^a=5^{2k+1} \equiv_8 5 \cdot 1
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  14. #14
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    It seems that Sudharaka's solution is incorrect. He claims that the equation 9x_1 + 16y_1=25z_1 has an integer solution, and while that claim may be correct, we have no idea that this solutions will be of the form (x_1,y_1,z_1) = (3^{x_0}, 4^{y_0}, 5^{z_0}) with x_0,y_0,z_0 integers...
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    Hi ... I've been disturbed by the proof by Sudharaka, because it somehow excludes the solution (0,1,1).

    So, I've been looking into it, and I spotted what I think is a problem.

    x', y', and z'are integers, so when you let x' = x_0 + 2, y' = y_0 + 2 and z' = z_0 + 2, then x_0, y_0, and z_0 can be negative (but not less than -2).

    This means that x_1, y_1, and z_1 are not necessarily integers, they may be fractions ... but what you have done is limited them to integer solutions ... I guess this means that your contradiction does not mean there are no other solutions to the given problem, but that 4^{x_0} can't be an integer ....

    Defunkt, I think you're thinking about it the wrong way ... indeed, what he has shown is that the solutions for 9x_1 + 16y_1=25z_1 cannot be of that form, hence no other solutions exist
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