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Question on the mod operator.

  1. Feb 7, 2006 #1
    If a + b = c, then is a + b = c (mod n) for all n?

    For example while reading LeVeque's Topics in Number Theory I came across a section on Fermat's Last Theorem in which he says: a way to show c^n = a^n + b^n has no solution is to assume the infinite amount of congruences c^n = a^n + b^n (mod p) for p = 2, 3, 5, 7, ... and then derive a contradiction.

    I assume what he means is: if c^n = a^n + b^n, then c^n = a^n + b^n (mod n) for any n that we choose.

    Is this valid?
  2. jcsd
  3. Feb 7, 2006 #2


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    What's the definition of equality, mod n?

    It's ambiguous.

    If you meant:

    For any n: [ if c^n = a^n + b^n, then c^n = a^n + b^n (mod n) ]

    That's fine. (Assuming you prove the theorem you originally asked)

    If you meant

    If c^n = a^n + b^n, then for any n: c^n = a^n + b^n (mod n)

    the no, that's invalid. n already has a meaning, so you cannot introduce n as a dummy variable! You'd have to use a new letter, like p.
  4. Feb 7, 2006 #3
    Sorry that was clumsy of me, I didn't realize that I was using n as the variable for the exponent already. What I mean to say is:
    If c^n = a^n + b^n, then under the mod equivalence c^n = a^n + b^n (mod k), where we can choose k at our leisure.

    It might be trivial but it seems to me that something is missing between these two steps.
  5. Feb 7, 2006 #4
    Yes, that's trivial. The definition of the mod k equivalence class is x = y (mod k) iff k|x-y. Thus c^n = a^n + b^n implies c^n-(a^n + b^n) = 0, which is divisible by k for any k (except 0).
  6. Feb 7, 2006 #5
    Ahh... of course that makes sense. I guess I just needed some reassurance to see it.

    So then by convention zero is divisible by any n > 0, thats very interesting!

    Thanks for the help.
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