The number of solutions of $2^{2x}-3^{2y}=55$, in which $x$,$y$ are integers
I tried using hit and trial and got only one solution $x=3$ and $y=1$ which is the answer.
How can we prove that it is the only solution?
2026-02-23 09:32:40.1771839160
The number of solutions of $2^{2x}-3^{2y}=55$, in which $x$,$y$ are integers
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1
$$ (2^x)^2-(3^y)^2=55 $$
$$ (2^x+3^y)(2^x-3^y)=11 \cdot 5 $$
Now two cases are possible.
$ 2^x+3^y=5, 2^x-3^y=11 $ case is eliminated. (Why?)
So $2^x+3^y=11, 2^x-3^y=5$
Adding them
$ 2 \cdot 2^x=16 $
Solving this you get $x=3$ and $y=1$