Saturday, May 13, 2017

elementary number theory - Proof : aequivbpmodnimpliesaiequivbipmodn




The congruence equation ab(modn)kZ,ab=kn.
Taking the i-th power of both sides : (ab)i=kini.




(ab)(ab)i1=kini.
n(ab),(ab)(ab)in(ab)imZ,(ab)i=mn.



But, my proof is incomplete, as it does not show still that aibi=kn.






All the comments and answer have ignored the fact that how to get the term aibi in the first place.
It seems that all have taken the approach to simply take the aibi=kn as a new equality, not derived from the original one. Just it uses the property of n(ab) of the original one.



Answer



It might be easier to use a=kn+b and the Binomial theorem. Then ai=(kn+b)i=ij=0(ij)(kn)jbij. All of these terms have a factor of n in them except for the 0th one, so this equation may be written in the form ai=pn+bi. Hence the claim.


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