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Since Cyclotomic polynomials are irreducible over $\mathbb{Q}$, $\phi_n(x)$, $\phi_m(x)$ are coprime as polynomials in $\mathbb{Z}[x]$.

  1. Working over $\mathbb{Q}$, $(\phi_n(x)$, $\phi_m(x))=(1)$. This implies that $(\phi_n(x), \phi_m(x))=(c)$ for some $c \in \mathbb{N}$ when this ideal is considered in $\mathbb{Z}[X]$. Can this $c$ be evaluated as a function of $n,m$?
  2. What can be said about $f(x) = gcd(\phi_n(x),\phi_m(x))$, when the polynomials are considered as scalars, i.e. evaluated at some $x$? $\forall x: f(x) | c$ from question 1, but can something stronger be said? More concretely, what is the image of $f$?

EDIT:

This paper by Apostol, provided in the comments to Greg's answer, gives a pretty good answer, which Greg guessed.

It basically calculates the resultant of 2 cyclotomic polynomials, which gives a number that is divisible by the optimal constant $c=(\phi_n(x), \phi_m(x))$, and it is either 1 (when $\frac{n}{m}$ is not a prime power) or a power of a prime ($p^{\phi(m)}$ when $\frac{n}{m}$ is a power of $p$).

When it is $1$ (the common case), we have a full answer for both questions. When it is a prime power, we only have an upper bound for the multiplicity ($\phi(m)$, when $m|n$).

This elementary paper gives some weaker result but it is simpler.

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Thanks anon, I removed the tag (originally I had a 3rd question about factoring Cyclotomic polynomials over finite fields but I found it on the site and removed it from the post). –  Ofir Dec 13 '12 at 8:10
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1 Answer

up vote 6 down vote accepted

The resultant $R$ of two polynomials $f,g$ has the property that there exist other polynomials $p,q$ such that $p(x)f(x)+q(x)g(x) = R$ identically. (Originally I had stated that $|R|$ is the least such positive integer, but this seems to be incorrect; see the comments.)

Therefore your question 1 is related to calculating the resultant of distinct cyclotomic polynomials $\phi_n, \phi_m$. Experimentally, the answer seems to be $1$ unless $m$ divides $n$ (or vice versa), in which case it seems to be a power of $n/m$. Just eyeballing some data, it seems the answer is $\exp(\phi(m)\Lambda(n/m))$, where $\Lambda$ is the von Mangoldt function.

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I've seen proofs of the exitence of $f,g$ to make $R$ but never a proof its the least. Can you give some sort of reference to a proof? –  dinoboy Dec 13 '12 at 1:30
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Greg, the resultant is not, in general, the least integer. See my paper, On resultants, Proc. Amer. Math. Soc. 89 (1983), 419-420, ams.org/journals/proc/1983-089-03/S0002-9939-1983-0715856-2/… –  Gerry Myerson Dec 13 '12 at 1:33
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See also Tom Apostol, Resultants of cyclotomic polynomials, Proc. Amer. Math. Soc. 24 (1970), 457-462, ams.org/journals/proc/1970-024-03/S0002-9939-1970-0251010-X/… –  Gerry Myerson Dec 13 '12 at 1:36
    
@GerryMyerson - Thank you for those papers, they fully answer my first question and they are readable too... –  Ofir Dec 13 '12 at 10:20
    
@GregMartin - Your empirical answer coincides with Apostol's paper. –  Ofir Dec 13 '12 at 10:23
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