Find all rational real numbers $x$ for which $\log_2(x^2 - 4x - 1)$ is a whole number.
The results are supposed to be: $x_1 = 5$, $x_2 = -1$, $x_3 = 17/4$, $x_4= -1/4$.
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What did you attempt? Hint: you need $x^2-4x-1=2^n$ for integer $n$. – egreg Nov 26 '16 at 11:16
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I tried to solve the equation in hope of getting a result. I'll try out what you wrote. Thanks, your help is much appreciated! – A M Nov 26 '16 at 11:19
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Can I use the quadratic formula here or do I need to change my approach? So my C would be c = -1 - 2^n – A M Nov 26 '16 at 11:21
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Distinguish between $n>0$ and $n<0$. The $n>0$ case gives $n=2$, so $x=5$ or $x=-1$. – egreg Nov 26 '16 at 11:34
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Put $y = x- 2 \implies x^2-4x-1 = 2^m \implies (x-2)^2 = 2^m+5\implies y^2 = 2^m+5$. This shows $y$ must be an integer since $2^m+5$ is an integer. If $m$ is odd, say $m = 2k+1\implies (y-1)(y+1) = 2(4^k+2)= 4(2^{2k-1}+1)$. Thus $y$ is odd. So put $y = 2r+1\implies 4r(r+1) = 4(2^{2k-1}+1)\implies r(r+1) = 2^{2k-1}+1$. This equation has no integer solution since the left side is even while the right side is odd. Thus $m$ must be even, then put $m = 2s\implies (y-2^s)(y+2^s) = 5$. I am sure you can take it from here...
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