October 21st, 2018, 05:07 AM  #1 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176  n=?
53(5^n 1), n=?

October 21st, 2018, 05:22 AM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry 
Have you heard of Fermat's little theorem?

October 21st, 2018, 06:20 AM  #3 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176 
so...!?

October 21st, 2018, 06:24 AM  #4 
Senior Member Joined: Oct 2009 Posts: 753 Thanks: 261 
Have you?

October 21st, 2018, 06:54 AM  #5 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176 
somewhat...

October 21st, 2018, 07:10 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,383 Thanks: 2011 
Have you considered n = 0?

October 21st, 2018, 07:14 AM  #7 
Senior Member Joined: Apr 2014 From: Europa Posts: 575 Thanks: 176 
n>0

October 21st, 2018, 02:55 PM  #8 
Member Joined: Oct 2018 From: Netherlands Posts: 39 Thanks: 3 
n=52 appears to be a solution but there are probably other ones. Didn't find it by trying but not through solid proof either. The values for nx26 are promising candidates by virtue of Fermat's little theorem, but they need to be verified as not all do the job. Last edited by skipjack; October 21st, 2018 at 04:54 PM. 
October 21st, 2018, 05:19 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,383 Thanks: 2011 
Any positive integer multiple of 52.

October 22nd, 2018, 02:47 AM  #10 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  Yes, that these numbers are solutions is immediate from Fermat's little theorem. The sightly more challenging thing is to show that there are no other solutions. But this problem can be reduced to showing that 4 and 26 are not solutions. Indeed, if n is a solution, then so is gcd(n,52) (by Bezout's lemma, for example). But if n is not a multiple of 52, then gcd(n,52) must divide 4 or 26 and so 4 or 26 would have to be a solution.
