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Hello today my teacher told us to find the solutions for $$n^2-2^m = 1$$ I came up with n= square root of 33 and m = 5, but after a review i saw that my calcultions were wrong ( i think ). could someone explain me if it was just a case or i did something right in my calculations ? thanks in advance.

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    $\begingroup$ Partial Differential Equations? $\endgroup$
    – lulu
    Oct 23, 2023 at 13:47
  • $\begingroup$ You calculated a correct pairing, but my guess is that $n,m$ are expected to be integers. I would start with a smaller $n$ to begin with... $\endgroup$
    – abiessu
    Oct 23, 2023 at 13:48
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    $\begingroup$ There's an obvious solution in integers, presumably this is what you were meant to look for. There are infinitely many solutions if you don't require them to be integers. $\endgroup$
    – lulu
    Oct 23, 2023 at 13:48
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    $\begingroup$ Does this answer your question? Find all answers of $n^2-2^m=1$ - found using an Approach0 search. As lulu's comment indicates, this is assuming that $m$ and $n$ are integers since, otherwise, there are infinitely many solutions. $\endgroup$ Oct 23, 2023 at 19:32
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    $\begingroup$ A nice little fact to carry around is Mihailescu's theorem, which states that the only two perfect powers (greater than $1$) that differ by $1$ are $8,9$ $\endgroup$ Oct 23, 2023 at 21:03

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You can rewrite that as $2^m = n^2 -1$ which is the same as $2^m = (n+1)\times(n-1)$

It immediately yields that both $n+1$ and $n-1$ are exact powers of 2. We can say that $\dfrac{n+1}{n-1}$ must be $2$: it cannot be 1 (would lead to $2 = 0$!), and any value strictly greater than 2 (4, 8, ...) would imply $n < 0$.

So the solution is given by: $$\dfrac{n+1}{n-1}=2 \Leftrightarrow n+1 = 2n -2 \Leftrightarrow n=3$$

We can deduce immediately that $2^m = 3^2 -1 = 8$, so $m=3$.

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