$n^2-2^m = 1$, cant find the answer

Question

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.

Answer 1

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 $m=3$.