I'm trying to solve this equation
$$ 2^n + 1 = k^2 $$ but I found only one solution where $$ n=k=3 $$ Is there are any other solutions and how to prove if there is no more?
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For $n$ even, there is no solution, because $2^n+1\equiv 1+1=2\not\equiv k^2\bmod 3$. – J. W. Tanner May 03 '20 at 15:35
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$2^n + 1 = k^2 \implies 2^n =k^2-1=(k+1)(k-1)\implies$ that $k+1$ and $k-1$ are both powers of $2$. – fleablood May 03 '20 at 16:12
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1So we have $k-1=2^m$ and $k+1 = 2^w$. But $2^w - 2^m = (k+1)-(k-1)=2$. Is $2^w-2^m = 2$ possible? If so... when? – fleablood May 03 '20 at 16:40
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@fleablood please put your answer and I'll accept it – nutella_eater May 03 '20 at 16:55
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Can't. Your questions been closed. And as pointed out it's already been answered. – fleablood May 03 '20 at 17:30