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Math Help - Diophantine equation

  1. #1
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    Diophantine equation

    Hi, I've to solve the following diophantine equation (in positive integers):

    3^x-5^y=z^2.

    Thanks for any help.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Arczi1984 View Post
    Hi, I've to solve the following diophantine equation (in positive integers):

    3^x-5^y=z^2.

    Thanks for any help.
    Maybe if we look the diophantine equation above \mathbb{Z}_4=\{0,1,2,3\}.

    It seems (only in theory!) that the above diophantine equation had no solution in positive integers...

    Last edited by Also sprach Zarathustra; July 6th 2010 at 11:58 PM.
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    It seems (only in theory!) that the above diophantine equation had no solution in positive integers...
    It has at least one solution, because 3^2-5^1 = 2^2.
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  4. #4
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    Quote Originally Posted by Arczi1984 View Post
    Hi, I've to solve the following diophantine equation (in positive integers):

    3^x-5^y=z^2.

    Thanks for any help.
    The solution is  (2,1,2)

    Take modulo  4 we have

     (-1)^x - 1 \equiv z^2 \bmod{4}

     x = 2t otherwise if  x  is odd we have  z^2 \equiv -2 \equiv 2 \bmod{4} which is impossible .


    Thus we have  3^{2t}  - 5^y = z^2

     (3^t)^2 - z^2 = 5^y

     (3^t - z )(3^t + z ) = 5^y which leads to

     3^t - z = 5^{\alpha} ~,~ 3^t  + z = 5^{\beta} where  \alpha + \beta = y


     3^t = \frac{ 5^{\beta} + 5^{\alpha} }{2}  which is the multiple of  5 if  \alpha \beta > 0 which is again impossible .

    Thus  \alpha = 0 ( not  \beta because  \alpha \leq \beta ) but they can't be both zero as we are looking for positive integers  y

     3^t = \frac{ 1 + 5^y }{2} ~~~~,~ \beta  = y-0 = y


    We now show that  (t,y) = (1,1) is the only solution .

    Suppose  t \geq 2 so we have

     \frac{ 1 + 5^y }{2} \equiv 0 \bmod{9} ~ \implies 1+ 5^y \equiv 0 \bmod{9}

    If we write down the first six powers of  5 modulo 9 , we obtain :  1,5,7,8,4,2,(1)

    we have exactly  5^6 \equiv 1 \bmod{9} and exactly  5^{3} \equiv -1 \bmod{9}

    We conclude that  y = 6k + 3 = 3(2k+1) for some non-negative integers  k


    Back to this equation :  3^t = \frac{ 1 + 5^y }{2}

    write  y = 3(2k+1) = 3m we have

     3^t = \frac{ (1+5^3)(1-5^3 + 5^6 - ... + 5^{3m-3} )}{2} = 9 \cdot 7P   the multiple of  7 which is also impossible .

    Therefore ,  0 < t \leq 1 ~ \implies t = 1

     t=1 is the only hope , luckily , we obtain  y = 1 and  z =2 ,  (2,1,2) is the only solution .
    Last edited by simplependulum; July 7th 2010 at 01:38 AM.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by simplependulum View Post
     t=1 is the only hope , luckily , we obtain  y = 1 and  z =2 ,  (2,1,2) is the only solution .
    Just in case you didn't follow how simplependulum obtained  y=1 , if  y>1 then  z^2=3^2-5^y<0 .
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