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Congruence in Z(integers) mod n

  1. Jan 22, 2009 #1
    1. The problem statement, all variables and given/known data

    In Z mod 5, compute (a + b)5.

    2. Relevant equations

    3. The attempt at a solution

    Noticing that (a+b)5 = a5+5(a4*b+2*a3*b2+2*a2*b3+a*b4)+b5. Since 5=0 in Z mod 5, it follows that 0(a4*b+2*a3*b2+2*a2*b3+a*b4)=0 and hence (a+b)5=a5+b5.

    I am just wondering if it is correct for me to say 5=0 in Z mod 5 and just substitute 0 in for 5?
  2. jcsd
  3. Jan 23, 2009 #2


    Staff: Mentor

    [tex]\equiv[/tex]Sounds reasonable to me. Why don't you test this with a couple of numbers to see if your results are consistent with what you've found?

    BTW, instead of saying 5=0 in Z mod 5, you can say 5 [itex]\equiv[/itex] 0 mod 5. That 3-bar equals sign means "is equivalent to".
  4. Jan 30, 2009 #3
    thanks for the response, much appreciated
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