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 December 26th, 2018, 12:35 PM #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, 02:39 PM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 327 Thanks: 42 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) Thanks from topsquark
 December 27th, 2018, 02:33 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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