0

I'm reading a book that claims that this is true. Is there a short/easy proof for this?

Michael Hardy
  • 1
Postskjerm
  • 445
  • It's true for all numbers, not only integers $a$. Can be realized looking on the complex unit circle for the roots of unity that they need to overlap and that will only happen when the mentioned condition occurs. – mathreadler May 03 '17 at 20:21
  • 1
    See also, for a more general theorem: https://math.stackexchange.com/questions/262130/how-to-prove-gcdam-bm-an-bn-a-gcdm-n-b-gcdm-n/262145 – Thomas Andrews May 03 '17 at 20:21
  • @mathreadler What do you mean "all numbers?" If $a=\frac{1}{2}$, what does it mean for divisibility? – Thomas Andrews May 03 '17 at 20:22
  • Please see the linked questions (yours is a duplicate of the first). Also be aware that it would have also been entirely appropriate to close your question as lacking any context whatsoever: you've no input or thoughts on the question, and no citation from the book your refer to. – amWhy May 03 '17 at 20:23
  • @amWhy Ok, I'l remember that. – Postskjerm May 03 '17 at 20:26
  • @ThomasAndrews Divisible in the sense that all roots in denominator overlap with those in numerator so that there are no roots in the denominator unaccounted for and we are sure to get a polynomial and not a rational function. – mathreadler May 03 '17 at 20:26
  • stenvik You see, I've reviewed three of your most recent question, and they all start "I'm reading a book...(that claims, that says......). Please name which book you are now referring to in this post. – amWhy May 03 '17 at 20:28
  • What roots, what denominator? You seem to be using terms very loosely. What if $a=\pi$. What does it mean that $\pi^r-1\mid \pi^n-1$? This is not a question about polynomials - if $p(x),q(x)$ are polynomials, it is possible that $p(x)\not\mid q(x)$ but $p(a)\mid q(a)$. @mathreadler – Thomas Andrews May 03 '17 at 20:33

0 Answers0