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: Southern California, USA Posts: 1,410 Thanks: 715 
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]?


Tags 
cardinality 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
cardinality  Roli  Probability and Statistics  1  June 18th, 2014 11:35 AM 
Cardinality of integers equals cardinality of reals  BenFRayfield  Number Theory  0  February 15th, 2014 02:55 PM 
Cardinality  arthurduh1  Real Analysis  11  October 21st, 2010 02:21 PM 
Cardinality  Mighty Mouse Jr  Algebra  8  October 19th, 2010 10:46 AM 
cardinality, ZFC  xboxlive89128  Applied Math  0  April 24th, 2010 04:39 PM 