2
$\begingroup$

It is known that the seventh coefficient of $\Phi_{105}(x)$ is $-2$ and that's the first occurrence of a coefficient with absolute value greater than $1$ for a cyclotomic polynomial. When I did a quick check for the seventh coefficient of $\Phi_n(x)$ where $n=105k$ with $\gcd(105,k)=1$ and $\mu(k)\neq 0$ they all came out to be $2$ in absolute value whenever they are nonzero. Is it true in general or there is a counterexample to this?

$\endgroup$
3
  • $\begingroup$ In a recent question of mine, "Cyclotomic polynomials: Φn(p) is like pϕ(n) for big enough p, right?" (See Related list for correct title and link), I was given a link to notes of Jameson. These notes spend some time on the coefficients and may help you with your answer. Gerhard "Thanks Again To Peter Mueller" Paseman, 2015.12.01 $\endgroup$ Dec 1, 2015 at 17:22
  • 1
    $\begingroup$ What cases did you check that worked? $\endgroup$ Dec 1, 2015 at 18:24
  • $\begingroup$ I guess I should've added the assumption $c_{7}\neq 0$ in which case it is true that $|c_{7}|=2$ as Ofir proved it. $\endgroup$ Dec 1, 2015 at 18:31

2 Answers 2

11
$\begingroup$

$k=11$ is the smallest counterexample - the 7'th coefficient is 0. Here are the details:

We have the following identity: $$\Phi_n(x) = \prod_{d \mid n} (1-x^d)^{\mu(n/d)},$$ valid for $n>1$.

If we are interested only in the first $m+1$ coefficients ($x^0$ to $x^{m}$), it suffices to look at the following product, going only over divisors $\le m$: $$\Phi_n(x) = \prod_{d \mid n, d \le m} (1-x^d)^{\mu(n/d)} \mod {x^{m+1}}.$$ Hence, $$[x^7] \Phi_{105 k}(x) = [x^7]\prod_{d \mid 105k, d \le 7} (1-x^d)^{\mu(105k/d)}.$$ Since you assume $\gcd(k,105)=1$ and $\mu(k)\neq 0$, we actually have 4 cases, according to the parity of $k$ and according to $\mu(k)$.

When $2 \nmid k$, the set $\{d : d\mid 105k, d \le 7\}$ is $\{1,3,5,7\}$ and we find $$[x^7] \Phi_{105 k}(x) = [x^7] (1-x)^{\mu(105k)}(1-x^3)^{\mu(35k)}(1-x^5)^{\mu(21k)}(1-x^7)^{\mu(15k)}$$ $$ = [x^7] ((1-x)^{-1}(1-x^3) (1-x^5)(1-x^7))^{\mu(k)}.$$ When $\mu(k)=1$, we get $-2$. When $\mu(k)=-1$, we get $0$.

When $2 \mid k$, the set $\{d : d\mid 105k, d \le 7\}$ is $\{1,2,3,5,6,7\}$ and we find $$[x^7] \Phi_{105 k}(x) = [x^7] (1-x)^{\mu(105k)}(1-x^2)^{\mu(105k/2)}(1-x^3)^{\mu(35k)}(1-x^5)^{\mu(21k)}(1-x^6)^{\mu(35k/2)}(1-x^7)^{\mu(15k)}$$ $$ = [x^7] ((1-x)(1-x^2)^{-1}(1-x^3)^{-1}(1-x^5)^{-1}(1-x^6)(1-x^7)^{-1})^{\mu(k/2)}.$$ Again, only two cases to check. When $\mu(k/2)=1$ we get 2, and when $\mu(k/2)=-1$ we get 0.

$\endgroup$
2
  • $\begingroup$ did you mean at the end"..., and when $\mu(k/2)=-1$ we get 0. $\endgroup$ Dec 1, 2015 at 18:20
  • $\begingroup$ @user204463 Yes, I had a slight miscalculation which I have corrected now. $\endgroup$ Dec 1, 2015 at 18:34
0
$\begingroup$

Actually for all $11 \leq k \leq 100$ which satisfy the conditions in the question (i.e. $\gcd(k,105) = 1$ and $\mu(k) \neq 0$), the coefficient of $x^7$ in $\Phi_{105k}(x)$ equals $0$. Hence so far there are only counterexamples!

This can be checked with GAP as follows:

gap> ks  := Filtered([1..100],k -> Gcd(k,105) = 1
>                              and Factors(k) = Set(Factors(k)));
[ 1, 2, 11, 13, 17, 19, 22, 23, 26, 29, 31, 34, 37, 38, 41, 43, 46, 47, 
  53, 58, 59, 61, 62, 67, 71, 73, 74, 79, 82, 83, 86, 89, 94, 97 ]
gap> c7s := List(ks,k->[k,CoefficientsOfUnivariatePolynomial(
>                           CyclotomicPolynomial(Rationals,105*k))[8]]);
[ [ 1, -2 ], [ 2, 2 ], [ 11, 0 ], [ 13, 0 ], [ 17, 0 ], [ 19, 0 ], 
  [ 22, 0 ], [ 23, 0 ], [ 26, 0 ], [ 29, 0 ], [ 31, 0 ], [ 34, 0 ],
  [ 37, 0 ], [ 38, 0 ], [ 41, 0 ], [ 43, 0 ], [ 46, 0 ], [ 47, 0 ],
  [ 53, 0 ], [ 58, 0 ], [ 59, 0 ], [ 61, 0 ], [ 62, 0 ], [ 67, 0 ],
  [ 71, 0 ], [ 73, 0 ], [ 74, 0 ], [ 79, 0 ], [ 82, 0 ], [ 83, 0 ],
  [ 86, 0 ], [ 89, 0 ], [ 94, 0 ], [ 97, 0 ] ]

(Note that the $8$ is not a typo, since the first list entry is the coefficient of $x^0$.)

$\endgroup$
1
  • 1
    $\begingroup$ yes it's because the odd ones are prime for which $\mu(k)=-1$ and the even ones are 2 times a prime in which case $\mu(k/2)=-1$ . For both cases @Ofir proved that the answer is 0. You have to go as high as 143 to see a $-2$ and as high as 286 to see a $2$. Other examples under 300 for a $-2$ are 187,209,221,247,253,299. If you double those numbers you will see a 2. $\endgroup$ Dec 2, 2015 at 19:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.