mersenneforum.org Solve phi(m^n+n)=2^n over positive integers?
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 2021-02-13, 12:59 #1 chao wu   Feb 2021 1 Posts Solve phi(m^n+n)=2^n over positive integers? How to solve phi(m^n+n)=2^n over positive integers? Where both m and n are positive integers, and phi denotes the Euler function. We can find that (m,n)=(2,1), (3,1) or (5,1) are some examples of solutions. Can someone give me any hints for this problem? MODERATOR NOTE: Moved to Homework Help. Last fiddled with by Dr Sardonicus on 2021-02-13 at 14:21 Reason: As indicated
 2021-02-13, 19:08 #2 Nick     Dec 2012 The Netherlands 110011100102 Posts You could start by thinking about which positive integers k have ϕ(k)=2ⁿ.
 2021-02-13, 19:53 #3 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 32×7×149 Posts This reminds me of the George Pólya book How to Solve It. Everyone could do very well by reading it. (Some time in their life, I mean.) Nick's suggestion fits the patterns Work backward, Eliminate possibilities, Consider special cases (or something like that, I am shooting from the hip).
 2021-02-13, 21:46 #4 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 2·727 Posts If x is composite then we know: c*x/log(log(x))0 is a constant [c=0.25 is good for all x>6]. ok, not very elegant to use these, though this is still elementary. With this you can easily solve the problem, the remaining x=m^n+n prime case is very easy. Last fiddled with by R. Gerbicz on 2021-02-13 at 21:47

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