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December 26th, 2018, 11:35 AM   #1
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congruence

how many solutions does this have for j and k
(x^71)-1≡ 0(mod (7^j)*(11^k))
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December 26th, 2018, 01:39 PM   #2
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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)
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December 27th, 2018, 01:33 AM   #3
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but how did you get x=1+(7^j)(11^k)
from (x^71)=1+(7^j)(11^k)w?
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