20210213, 12:59  #1 
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 20210213 at 14:21 Reason: As indicated 
20210213, 19:08  #2 
Dec 2012
The Netherlands
11001010000_{2} 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
246C_{16} 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
2×7×103 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 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Sums of a positive cube and a power of 2  enzocreti  enzocreti  6  20200219 04:47 
Fun with a false positive  Madpoo  Data  12  20160629 19:00 
another false positive?  ixfd64  Data  3  20160314 22:11 
positive LL test?  sixblueboxes  PrimeNet  90  20140724 05:51 
False positive?  Pi Rho  Lounge  4  20030423 14:11 