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.
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.
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?