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Math Help - Equation with exponents

  1. #1
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    Equation with exponents

    Could you help me solve the following equation:

    2^x + 5^y = k^2

    where x, y, k are integers.

    I know the only solutions are (3,0,3) and (2,1,3). The problem is - I don't know how to prove it. Please help.
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  2. #2
    Member sbhatnagar's Avatar
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    Re: Equation with exponents

    I understand your problem well. You can find an expression for x,y and k.







    Similarly,



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  3. #3
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    Re: Equation with exponents

    Thanks, but could you tell me where do those three equations lead to?
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  4. #4
    Member sbhatnagar's Avatar
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    Re: Equation with exponents

    y=0,1
    k=3
    Therefore x=ln(9-5)/ln(2)=2
    And x=ln(9-1)/ln(2)=3

    Similarly find the values for y and k. But remenber you cant find the values of three variables from just 1 equation.
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  5. #5
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    Re: Equation with exponents

    Thank you very much for the hint. Could you please take a quick look at what I came up with as I was waiting for your reply?

    If y=0, we have 2^x=k^2-1=(k+1)(k-1), so k+1 and k-1 are powers of 2, and the only powers of 2 that differ by 2 are 2 and 4, so (x,y,k)=(3,0,3).

    If y>0, 2^x is a quadratic residue modulo 5, so x must be even. Hence 5^y=k^2-2^x=(k+2^{0,5x})(k-2^{0,5x}). The factors are powers of 5, say k+2^{0,5x}=5^m and k-2^{0,5x}=5^n.

    We now have 5^m-5^n=2^{0,5x+1}. Since the right side is not divisible by 5, we must have n=0(is it correct?). Then, for m=1, we get x=2, and so (x,y,k)=(2,1,3). But what next? How do I prove that there are no more solutions?
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