20210213, 12:59  #1 
Feb 2021
1_{8} 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 20210213 at 14:21 Reason: As indicated 
20210213, 19:08  #2 
Dec 2012
The Netherlands
11×151 Posts 
You could start by thinking about which positive integers k have ϕ(k)=2ⁿ.

20210213, 19:53  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,391 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). 
20210213, 21:46  #4 
"Robert Gerbicz"
Oct 2005
Hungary
1,459 Posts 
If x is composite then we know:
c*x/log(log(x))<phi(x)<=xsqrt(x), where c>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 20210213 at 21:47 
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