Does anybody know the all positive integer solution of $$ a^x-b^y=2 $$ under the condition $$ x \geq 2, y\geq2 $$
I didn't find any solution without $$ x=5,y=3,a=2,b=3 $$
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$3^3-5^2=2$ is one. – Bennett Gardiner Sep 06 '15 at 01:30
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Related: Solutions to $7^a+2=3^b$, Solutions to $x^{2}-7^{y}=2$. – Bart Michels Sep 06 '15 at 11:30
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See Catalan's conjecture (Mihăilescu's theorem) for $a^x-b^y=1$, where it's also written about the general $a^x-b^y=n$.
https://oeis.org/A076427 says there's exactly one solution when $a^x,b^y<10^{18}$.
The solution is $(a,b,x,y)=(3,5,3,2)$.
user236182
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This is the only known solution. It is an open problem as to whether it is the only solution. – Mike Bennett Sep 06 '15 at 16:05