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April 20th, 2017, 09:40 AM   #1
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Cardinality

How many k satisfy $\displaystyle \;$$\displaystyle 2^n -n=k^2$
Example , $\displaystyle 2^7 -7=11^2$ $\displaystyle \; $$\displaystyle n,k \in N$

Last edited by idontknow; April 20th, 2017 at 09:44 AM.
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April 20th, 2017, 11:22 AM   #2
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I can't prove it yet but it certainly looks like $(n,k)=(7,11)$ is the only integer pair such that

$2^n-n=k^2$
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April 21st, 2017, 12:42 AM   #3
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What about (n,k) = (1,1) or (n,k) = (0,1) [if you allow n = 0]?
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