How can the expression (a^n+b^n)/(a+b) be simplified?

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Discussion Overview

The discussion revolves around the simplification of the expression (a^n + b^n) / (a + b), exploring various cases based on the value of n and the nature of a and b. Participants consider different mathematical contexts, including modular arithmetic and polynomial division.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant queries whether the expression can be simplified or if it is already in its simplest form.
  • Another participant suggests that if n is prime and a, b are elements of Z/nZ, the expression simplifies to (a + b)^(n-1).
  • A different viewpoint indicates that if n is even, the expression cannot be simplified, while if n is odd, long division can be performed to yield a polynomial of lower degree.
  • Concerns are raised about the applicability of the prime condition and modular arithmetic to the original question, suggesting it may confuse the original poster (OP).
  • Some participants seek clarification on the meaning of Z/nZ and its relevance to the discussion.
  • There is a mention of the "freshman's dream" in modular arithmetic, which relates to the simplification of expressions involving powers.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which the expression can be simplified, particularly regarding the nature of n (prime, even, odd) and the context of a and b. There is no consensus on a single approach or solution.

Contextual Notes

Limitations include the lack of clarity on the definitions of a and b, the specific conditions under which simplifications apply, and the potential confusion arising from references to modular arithmetic.

eddybob123
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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)
 
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If n is prime and a,b are elements of Z/nZ then that is (a + b)^(n-1)
 
eddybob123 said:
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)

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

[tex]\sum_{k=0}^{n-1}a^{n-k-1}.(-b)^k[/tex]

Of course, I wouldn't call the latter form "more simplified". The greatest exponent is lower, though.
 
Last edited:
@Mandlebra
n should be prime but what exactly do you mean by "Z/nZ. As you can probably see, I am an amateur mathematician.
 
Znz is the ring of integers mod n. 0,1,2,..., n-1. Check out modular arithmetic on wiki for an idea
 
Does this work for all values of a and b (positive, negative, integral, non-integral, etc.)?
 
Mandlebra said:
If n is prime and a,b are elements of Z/nZ then that is (a + b)^(n-1)

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.
 
micromass said:
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.

Who knows. He didn't provide much info
 
What was the "Z/nZ" you were talking about in your previous post？
 
  • #11
eddybob123 said:
What was the "Z/nZ" you were talking about in your previous post？

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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