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Observe the following equations:

$2x^2 + 1 = 3^n$ has two solutions $(1, 1) ~\text{and}~ (2, 2)$

$x^2 + 1 = 2 \cdot 5^n$ has two solutions $(3, 1) ~\text{and}~ (7, 2)$

$7x^2 + 11= 2 \cdot 3^n$ has two solutions $(1, 2) ~\text{and}~ (1169, 14)$

$x^2 + 3 = 4 \cdot 7^n$ has two solutions $(5, 1) ~\text{and}~ (37, 3)$

How one can determine the only number of solutions are two or three or four...depends up on the equation. especially, the above equations has only two solutions. How can we prove there is no other solutions? Or how can we get solutions by any particular method or approach?

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    @pedja! waiting for a solution2012-03-26
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    [Waiting for Godot](http://en.wikipedia.org/wiki/Waiting_for_Godot)2012-03-26
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    [related](http://math.stackexchange.com/q/122846/19341)?2012-03-26
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    [this](http://math.stackexchange.com/q/118941/19341) too2012-03-26
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    @pedja! you have given a good material on "Waiting for Godot", really great.Thank you.2012-03-26
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    $(11, 5)$ for $(x,n)$ is also a solution of $2x^2 + 1 = 3^n$2012-03-26
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    It would help if you would let us know where you found these equations, and why you think they have only the solutions you have given.2012-03-26

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