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Simplify this expression

  1. Jun 15, 2012 #1
    Hi I was wondering if there is any way to simplify this expression or if it's already in its simplest form. Thank you in advance: (a^n+b^n)/(a+b)
  2. jcsd
  3. Jun 15, 2012 #2
    If n is prime and a,b are elements of Z/nZ then that is (a + b)^(n-1)
  4. Jun 15, 2012 #3


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    If n is even, it cannot be simplified.

    If n is odd, you can do long division without a remainder to get the following expression:

    [tex]a^{n-1} - ba^{n-2} + b^2a^{n-3} - .... + b^{n-1}[/tex]

    which is


    Of course, I wouldn't call the latter form "more simplified". The greatest exponent is lower, though.
    Last edited: Jun 15, 2012
  5. Jun 16, 2012 #4
    n should be prime but what exactly do you mean by "Z/nZ. As you can probably see, I am an amateur mathematician.
  6. Jun 16, 2012 #5
    Znz is the ring of integers mod n. 0,1,2,..., n-1. Check out modular arithmetic on wiki for an idea
  7. Jun 16, 2012 #6
    Does this work for all values of a and b (positive, negative, integral, non-integral, etc.)?
  8. Jun 17, 2012 #7
    I don't think the OP was talking about [itex]\mathbb{Z}/n\mathbb{Z}[/itex]. So I don't think it's good to post this kind of information since it only confuses the OP.

    OP: the answer you want has been given by curiouspi.
  9. Jun 17, 2012 #8
    Who knows. He didn't provide much info
  10. Jun 17, 2012 #9
    What was the "Z/nZ" you were talking about in your previous post?
  11. Jun 17, 2012 #10
    Last edited: Jun 17, 2012
  12. Jun 17, 2012 #11
    The numbers on a clock are Z/12Z, for example. 7+8=3 mod 12. What time is it 8 hours after 7:00?

    You don't really need these numbers till you take abstract algebra (AKA advanced algebra AKA modern algebra AKA algebra, not too be confused with those of the same name for high school, or linear algebra), or number theory or maybe combinatorics.

    But, if you do, then you learn that in Z/pZ, where p is prime, then you have "the freshman's dream" (a+b)^p=a^+b^p, which is based off a quick application of the binomial theorem and applying multiples of p in Z/pZ are 0 (mod p).
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