December 26th, 2018, 11:35 AM  #1 
Newbie Joined: Dec 2018 From: israel Posts: 5 Thanks: 0  congruence
how many solutions does this have for j and k (x^71)1≡ 0(mod (7^j)*(11^k)) 
December 26th, 2018, 01:39 PM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 647 Thanks: 94 
By inspection , set $\displaystyle x=1+ 7^j 11^k $ $\displaystyle 7^j 11^k \equiv 0 \; mod (7^j 11^k)$ So you have infinite pairs (j,k) 
December 27th, 2018, 01:33 AM  #3 
Newbie Joined: Dec 2018 From: israel Posts: 5 Thanks: 0 
but how did you get x=1+(7^j)(11^k) from (x^71)=1+(7^j)(11^k)w? 

Tags 
congruence 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Congruence in F[x]  Alexis87  Abstract Algebra  1  March 31st, 2018 09:25 AM 
Congruence  s.a.sajib  Number Theory  9  April 8th, 2013 06:28 AM 
congruence  justjones  Algebra  2  March 13th, 2013 02:38 PM 
Congruence  Fernando  Number Theory  3  May 15th, 2012 08:20 AM 
Congruence  Aizen  Number Theory  0  June 28th, 2008 12:05 PM 