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: iPhone Posts: 444 Thanks: 68 
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? 

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