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Relative Primes and mod

  1. Feb 26, 2009 #1
    Hi all,
    Supppose that n > 0 and 0 < x < n are integers and x is relatively prime to n, show that there is an integer y with the property:
    x*y is congruent to 1 (mod n)

    I have attempted the following, I am not sure if I am on the right track:
    1 = xy + qn which implies 1 - xy = qn
    n|(1-xy) which implies q(1-xy) = n
    so if I divide q in the first equation i get [tex]\frac{1-xy}{q}[/tex]=n which is equal to q(1-xy)=n.

    Thanks in advance
    Maunil
     
    Last edited: Feb 26, 2009
  2. jcsd
  3. Feb 27, 2009 #2
    There you go!
    1-xy=qn implies 1-xy=0 (mod n), hence 1=xy (mod n).

    No, n|(1-xy) implies (1-xy)=kn for some k, which you already knew.
     
  4. Feb 28, 2009 #3
    Another way to look at this is to remember that the elements of a residue system are always less than N.

    If we look at the powers of X they must come to repeat and [tex] X^S \equiv X^T Mod N [/tex] Since X is relatively prime to N we can cancel out powers of X.

    This gives [tex]X^{S-T} \equiv 1 Mod N [/tex] Assuming S is the larger value, we can not have S-T = 1 unless X is its own inverse.

    Thus S-T = 2 or more, and [tex] X^{S-T-1} \equiv X^{-1} Mod N. [/tex]
     
    Last edited: Feb 28, 2009
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