ab equals to cd mod nprove that whenever ab≡cd mod n and b≡d mod n with gcd(b,n)=1 then a≡c mod n.
steps by steps please because i just learning it. For gcd(b,n)=1,there exist two integer s and t such that sb +tn=1, orsb=1-tn,so sb=1 (mod n).Since b=d(mod n), we havesd=sb(mod n)=1(mod n) ……(*) For ab≡cd (mod n),(ab)s=(cd)s (mod n),= > a(bs)=c(ds) (mod n)(since (*) sd=sb =1 (mod n))= > a=c (mod n)[[Done]]
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ab equals to cd mod nprove that whenever ab≡cd mod n and b≡d mod n with gcd(b,n)=1 then a≡c mod n.
steps by steps please because i just learning it. For gcd(b,n)=1,there exist two integer s and t such that sb +tn=1, orsb=1-tn,so sb=1 (mod n).Since b=d(mod n), we havesd=sb(mod n)=1(mod n) ……(*) For ab≡cd (mod n),(ab)s=(cd)s (mod n),= > a(bs)=c(ds) (mod n)(since (*) sd=sb =1 (mod n))= > a=c (mod n)[[Done]]