April 20th, 2017, 09:40 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 154 Thanks: 21  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. 
April 20th, 2017, 11:22 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,301 Thanks: 664 
I can't prove it yet but it certainly looks like $(n,k)=(7,11)$ is the only integer pair such that $2^nn=k^2$ 
April 21st, 2017, 12:42 AM  #3 
Senior Member Joined: Nov 2010 From: Berkeley, CA Posts: 174 Thanks: 35 Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT 
What about (n,k) = (1,1) or (n,k) = (0,1) [if you allow n = 0]?


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