# Equation with exponents

• Sep 18th 2011, 03:54 AM
gollum
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.
• Sep 18th 2011, 04:21 AM
sbhatnagar
Re: Equation with exponents
• Sep 19th 2011, 07:55 AM
gollum
Re: Equation with exponents
Thanks, but could you tell me where do those three equations lead to?
• Sep 19th 2011, 08:44 AM
sbhatnagar
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.
• Sep 19th 2011, 09:29 AM
gollum
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?