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How one expect the possible values of $(x, y, z) = (0, 0,0), (1, 1, 1)$ and $(3, 1, 5)$ of the equation $3^x -2^y = z^2$ without by inspection.

Why $n = 1$ and $3$ are valid for $5^n - 4 = z^2$.

Is there any other $n$, other than these $1$ and $3$? without inspection how we prove the solutions/how we will find other solutions if there?

Prema

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    Google search "Ramanujan-Nagell equation" and you should find your answer2012-03-21
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    @prema: For $5^n-4=z^2$, I think I posted a solution on this site less than a month ago.2012-03-21
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    @Andre Was that "Solve $x^2+4=y^d$ for $d \geq 3$ question. It is http://tinyurl.com/7mchmtt (this one)2012-03-21
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    @KV Raman: My memory was not quite accurate. The question about $5^n$ is different, easier but different.2012-03-21
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    @Prema Is there a specific goal you have? Are you doing research? I would suggest you read a paper written by J.H.E.Cohn at http://tinyurl.com/7dgok6k. I changed the title so that more people could pay attention. I wish I knew the answer.2012-03-22
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    @prema I've been working through the first equation, and I've already found another solution - $3^4-2^5=7^2$. I'm in the middle of writing a post explaining the progress I've made; I suppose that this provides some justification for your question - given that inspection failed to find that solution, we have to find a better way!2012-03-24
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    @prema Another (easy) solution - $3^2-2^3=1^2$. In fact, it was this identity which allowed me to find the solution I mentioned in my previous comment.2012-03-24
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    (If $2^a+1=3^k$, then we can use the identity $(2^a+1)^2-4 \times 2^a = (2^a-1)^2$ to generate a solution, setting $x=2k$, $y=a+2$ and $z=(2^a+1)$)2012-03-24
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    That should say '$z=(2^a-1)$'.2012-03-24

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