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anonymous
 one year ago
Is this problem an application of the cancelation law of congruences? Prove that if bd ≡ bd' (mod p), where p is a prime, and p does not divide b, then d ≡ d' (mod p)
anonymous
 one year ago
Is this problem an application of the cancelation law of congruences? Prove that if bd ≡ bd' (mod p), where p is a prime, and p does not divide b, then d ≡ d' (mod p)

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1That follows trivially from euclid lemma : \(p\mid mn \implies p\mid m~~\text{or}~~p\mid n\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\[bd\equiv bd'\pmod{p} \implies p\mid b (dd')\] since \(p\nmid b\), it must be the case that \(p\mid (dd')\) which is equivalent to saying \(d\equiv d'\pmod{p}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah.... I see. Is gcd(b,p) = 1 though?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1\(p\nmid b ~~\iff~~\gcd(p,b)=1 \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so in general, m does not divide n if and only if gcd(m,n)=1?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1that is correct only if \(m\) is a prime

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1consider an example : \(m = 4,~ n = 6\) \(4\nmid 6\) but \(\gcd(4,6)\ne 1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1btw, in this thread \(p\) is assumed to be a prime.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah... I see. Thank you :')

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So there are two ways to do the problem above. Euclid lemma or cancelation law (since gcd(b,p) = 1)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1pretty sure there are many other ways to prove it, as you can see the proof is trivial
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