I think it must be finite, $y$ is always even, but I don't know how to continue.
edit: with $x,y\in\mathbb Z$
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Just by solving for $x$ or $y$ you can see that the solution set isn't finite. It's uncountably infinite. – zibadawa timmy Nov 19 '13 at 23:26
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4I guess he means in integers. – M.B. Nov 19 '13 at 23:26
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1I'm pretty sure $x$ has to be even too, but I don't have anything further than that. – Dennis Meng Nov 19 '13 at 23:34
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$x$ has to be even from considering mod 4. – M.B. Nov 19 '13 at 23:34
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By inspection, there is the "obvious" solution of $x=2$ and $y=-8$ or $y=8$. – Christopher K Nov 19 '13 at 23:34
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6$\pmod{4}$ gives $x$ even. Now write $$55=(y+3^{\frac{x}{2}})(y-3^{\frac{x}{2}})$$ – Ivan Loh Nov 19 '13 at 23:35
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1$x=6$, $y=28$ is the next solution. – rogerl Nov 19 '13 at 23:40
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ok if y is even then its last digit can be 0,4 or 6. and if x is odd then the last digit is either 3 or 7 which is not possible, but how is the even case – derivative Nov 19 '13 at 23:44
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1(fix) As quick observations, you have $y\equiv 0\bmod2$, $y^2\equiv1\bmod3$ (equivalently $y\not\equiv 0 \bmod 3$), $y\equiv\pm 1\bmod5$. $3^x+55\equiv(−1)^2x+3\equiv0\bmod4$, which forces x even as Dennis says. – zibadawa timmy Nov 19 '13 at 23:50
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Hint Modulo $4$ we have
$$(-1)^x \equiv 1 \pmod{4} \Rightarrow x =2k$$
Then
$$(y-3^k)(y+3^k) =55$$
Now all you have to do is check all possible factorizations of $55: 1 \times 55$ or $5 \times 11$.
N. S.
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