Could you help me solve the following equation:

$\displaystyle 2^x + 5^y = k^2$

where $\displaystyle x, y, k$ 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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- Sep 18th 2011, 03:54 AMgollumEquation with exponents
Could you help me solve the following equation:

$\displaystyle 2^x + 5^y = k^2$

where $\displaystyle x, y, k$ 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 AMsbhatnagarRe: Equation with exponents
I understand your problem well. You can find an expression for x,y and k.

http://latex.codecogs.com/gif.latex?2^x+5^y=k^2

http://latex.codecogs.com/gif.latex?2^x =k^2 - 5^y

http://latex.codecogs.com/gif.latex?...5^y)}}{\ln(2)}

Similarly,

http://latex.codecogs.com/gif.latex?...2^x)}}{\ln(5)}

http://latex.codecogs.com/gif.latex?k=\sqrt{2^x+5^y} - Sep 19th 2011, 07:55 AMgollumRe: Equation with exponents
Thanks, but could you tell me where do those three equations lead to?

- Sep 19th 2011, 08:44 AMsbhatnagarRe: 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 AMgollumRe: 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 $\displaystyle 2^x=k^2-1=(k+1)(k-1)$, so $\displaystyle k+1$ and $\displaystyle k-1$ 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, $\displaystyle 2^x$ is a quadratic residue modulo 5, so x must be even. Hence $\displaystyle 5^y=k^2-2^x=(k+2^{0,5x})(k-2^{0,5x})$. The factors are powers of 5, say $\displaystyle k+2^{0,5x}=5^m$ and $\displaystyle k-2^{0,5x}=5^n$.

We now have $\displaystyle 5^m-5^n=2^{0,5x+1}$. 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?