Dividing polynomials with coefficients

Question

Let $\;f(t)=\left(t^{pq}-1\right)\left(t-1\right)\;$ and $\;g(t)=\left(t^p-1\right)\left(t^q-1\right)\;$ where $p$ and $q$ are relatively prime positive integers.
Prove that $\dfrac{f(t)}{g(t)}$ can be written as a polynomial where it has just $1$ or $-1$ as coefficients. (For example, when $p=2$ and $q=3$, we have that $\dfrac{f(t)}{g(t)}=t^2-t+1$) .

After expanding the terms, the final division becomes $\dfrac{f(t)}{g(t)}=\dfrac{t^{p(q-1)+1}\!-\!t^{p(q-1)}\!+\!t^{p(q-2)+1}\!-\!t^{p(q-2)}\!+\!\ldots\!+\!t^{p+1}\!-\!t^p\!+\!t\!-\!1}{t^{q}-1}$

But with this result, it is still confusing how to prove coefficients are either $-1$ or $1$.

Answer 1

A possible approach is the following.

We split the division into two parts $$\frac{\sum_{i = 0}^{q - 1}x^{pi + 1}}{x^q - 1} - \frac{\sum_{i = 0}^{q - 1}x^{pi}}{x^q - 1}.$$ For the first part, note that $$\frac{x^{pi + 1}}{x^{q} - 1} = \sum_{a < pi + 1, a \equiv pi + 1 \bmod{q}} x^a + \frac{x^{(pi + 1) \% q}}{x^q - 1}$$ where $(pi + 1) \% q$ is the remainder of $pi + 1$ modulo $q$.

Summing over all $i$, we conclude that $$\frac{\sum_{i = 0}^{q - 1}x^{pi + 1}}{x^q - 1} = \sum_{i = 0}^{q - 1}\sum_{a < pi + 1, a \equiv pi + 1 \bmod{q}} x^a + \sum_{i = 0}^{q - 1} \frac{x^{(pi + 1) \% q}}{x^q - 1}.$$ The crucial observation is that, since $(p, q) = 1$, the numbers $\{pi: i = 0, \cdots, q - 1\}$ form a complete residual class modulo $q$. So we have $$\sum_{i = 0}^{q - 1} \frac{x^{(pi + 1) \% q}}{x^q - 1} = \frac{1 + x + \cdots + x^{q - 1}}{x^q - 1} = \frac{1}{x - 1}.$$ Furthermore, for any integer $a$, there is at most one solution to $a <pi + 1, a \equiv pi + 1 \bmod{q}$ in $i \in \{0, 1, \cdots, q - 1\}$. So each monomial appears in $$\sum_{i = 0}^{q - 1}\sum_{a < pi + 1, a \equiv pi + 1 \bmod{q}} x^a$$ at most one time. In other words, this term is a polynomial with coefficient in $\{0, 1\}$.

Similarly, $$\frac{\sum_{i = 0}^{q - 1}x^{pi}}{x^q - 1} = \sum_{i = 0}^{q - 1}\sum_{a < pi, a \equiv pi \bmod{q}} x^a + \sum_{i = 0}^{q - 1} \frac{x^{pi \% q}}{x^q - 1}.$$ The second term is $\frac{1}{x - 1}$ by the same reasoning as above. So we conclude that $$\frac{\sum_{i = 0}^{q - 1}x^{pi + 1}}{x^q - 1} - \frac{\sum_{i = 0}^{q - 1}x^{pi}}{x^q - 1} = \sum_{i = 0}^{q - 1}\sum_{a < pi + 1, a \equiv pi + 1 \bmod{q}} x^a - \sum_{i = 0}^{q - 1}\sum_{a < pi, a \equiv pi \bmod{q}} x^a.$$ As we have argued, both sums are a polynomial with coefficient in $\{0, 1\}$. So their difference must be a polynomial with coefficients in $\{-1,0,1\}$, as desired.

Answer 2

I'm going to prove the following two claims separately.

Claim 1 : $\dfrac{f(t)}{g(t)}$ is a polynomial.

Claim 2 : $\dfrac{f(t)}{g(t)}$ has only $0,\pm 1$ as the coefficients.

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Claim 1 : $\dfrac{f(t)}{g(t)}$ is a polynomial.

Proof :

We have $$\begin{align}\frac{f(t)}{g(t)}&=\frac{\left(t^{pq}-1\right)\left(t-1\right)}{\left(t^p-1\right)\left(t^q-1\right)} \\\\&=(t-1)\cdot\frac{t^{pq}-1}{t^q-1}\cdot\frac{1}{t^p-1} \\\\&=\frac{(t-1)(1+t^q+\cdots +t^{(p-1)q})}{t^p-1}\tag1\end{align}$$

Here, $t_m=\cos\frac{2\pi m}{p}+i\sin\frac{2\pi m}{p}\ (m=0,1,\cdots, p-1)$ are the roots of the denominator.

Every root of the denominator is a root with multiplicity one.

So, in order to prove that $(1)$ is a polynomial, it is sufficient to prove that every root of the denominator is a root of the numerator.

We have $t_0=1$. So, we want to prove that for every $m=1,2,\cdots, p-1$, $$1+(t_m)^q+(t_m)^{2q}+\cdots +(t_m)^{(p-1)q}=0$$

We have $$\begin{align}&1+(t_m)^q+(t_m)^{2q}+\cdots +(t_m)^{(p-1)q} \\\\&=1+\bigg(\cos\frac{2\pi mq}{p}+i\sin\frac{2\pi mq}{p}\bigg)+\bigg(\cos\frac{2\pi m\times 2q}{p}+i\sin\frac{2\pi m\times 2q}{p}\bigg) \\&\qquad+\cdots +\bigg(\cos\frac{2\pi m(p-1)q}{p}+i\sin\frac{2\pi m(p-1)q}{p}\bigg) \\\\&=1+\bigg(\sum_{k=1}^{p-1}\cos\frac{2\pi kmq}{p}\bigg)+i\sum_{k=1}^{p-1}\sin\frac{2\pi kmq}{p} \\\\&=\bigg(\sum_{k=0}^{p-1}\cos\frac{2\pi kmq}{p}\bigg)+i\sum_{k=0}^{p-1}\sin\frac{2\pi kmq}{p}\tag2\end{align}$$

In general, we have $$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$ and $$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2 a + (n-1)\cdot d}{2}\biggr)$$

(See here.)

Setting $n=p,a=0$ and $d=\frac{2\pi mq}{p}$, we have $$\sum_{k=0}^{p-1}\cos\frac{2\pi kmq}{p}=\frac{\overbrace{\sin(mq\pi)}^{=0}}{\sin ( \frac{mq\pi}{p} )} \times \cos \biggl( \frac{ (p-1)mq\pi}{p}\biggr)=0$$ and $$\sum_{k=0}^{p-1}\sin\frac{2\pi kmq}{p}=\frac{\overbrace{\sin(mq\pi)}^{=0}}{\sin ( \frac{mq\pi}{p} )} \times \sin\biggl( \frac{(p-1)mq\pi}{p}\biggr)=0$$

Therefore, $(2)=0$ follows.$\ \blacksquare$

(Some comments : I used $\gcd(p,q)=1$ in the above proof. Here, $\sin(mq\pi/p)$ has to be non-zero. If $\gcd(p,q)=1$, then $\sin(mq\pi/p)$ is non-zero for every $m=1,2,\cdots,p−1$.)

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Claim 2 : $\dfrac{f(t)}{g(t)}$ has only $0,\pm 1$ as the coefficients.

Proof :

For $t$ such that $|t|^p\lt 1$, we have $$\begin{align}\frac{f(t)}{g(t)}&=\frac{\left(t^{pq}-1\right)\left(t-1\right)}{\left(t^p-1\right)\left(t^q-1\right)} \\\\&=(1-t)\cdot\frac{t^{pq}-1}{t^q-1}\cdot\frac{1}{1-t^p} \\\\&=(1-t)(1+t^q+\cdots +t^{(p-1)q})(1+t^p+t^{2p}+\cdots) \\\\&=(1+t^q+\cdots +t^{(p-1)q})(1+t^p+t^{2p}+\cdots) \\&\qquad -t(1+t^q+\cdots +t^{(p-1)q})(1+t^p+t^{2p}+\cdots) \\\\&=1+t^p+t^{2p}+\cdots \\&\qquad +t^q+t^{p+q}+t^{2p+q}+\cdots \\&\qquad +\cdots +t^{(p-1)q}+t^{p+(p-1)q}+t^{2p+(p-1)q}+\cdots \\&\qquad -t-t^{p+1}-t^{2p+1}-\cdots \\&\qquad -t^{q+1}-t^{p+q+1}-t^{2p+q+1}-\cdots \\&\qquad -\cdots -t^{(p-1)q+1}-t^{p+(p-1)q+1}-t^{2p+(p-1)q+1}-\cdots\end{align}$$ Each term is of the form $(-1)^{a}t^{a+bp+cq}$ where $a,b,c$ are integers satisfying $$a=0,1,\quad b\ge 0,\quad 0\le c\le p-1$$If $a+bp+cq=a'+b'p+c'q$ with $a=a'$, then we have $(b-b')p=(c'-c)q$ which implies $p\mid (c'-c)$ because of $\gcd(p,q)=1$. Since $-p+1\le c'-c\le p-1$, we have $c'=c$ and $b=b'$.

If $a+bp+cq=a'+b'p+c'q$ with $a\not =a'$, then $(-1)^at^{a+bp+cq}+(-1)^{a'}t^{a'+b'p+c'q}=((-1)^a+(-1)^{a'})t^{a+bp+cq}=0$.

It follows from the identity theorem for power series (if two power series take the same values on an interval, then they are identical) that each coefficient of $\frac{f(t)}{g(t)}$ can equal only $0$ or $\pm 1$.$\ \blacksquare$

(some comments : Polynomials are power series (all polynomials are (finite) power series, not all power series are polynomials), and the coefficients of convergent power series are uniquely determined. From claim 1, we can write $\dfrac{f(t)}{g(t)}=\displaystyle\sum_{n=0}^{(p-1)(q-1)}a_nx^n$ where $a_{(p-1)(q-1)}=1$. We've seen that for $|t|^p\lt 1$, we can write it as $\displaystyle\sum_{n=0}^{\infty}b_nx^n$ with $b_n=0,\pm 1$. Then, from the identity theorem, we have $a_n=b_n=0,\pm 1$ for $0\le n\le (p-1)(q-1)$, and $b_n=0$ for $n>(p-1)(q-1)$.)

Answer 3

This is largely a rewrite of mathlove's proof. But I thought it was better to write it from scratch than modify his answer.

(1) Let's first show that $$\phi(t)=\frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)}$$ is a polynomial.

Let $\omega=\exp(2\pi i/pq)$, which is an order $pq$ root of unity: ie, $\omega^{pq}=1$, but no lower power of $\omega$ is one. This allows us to factor $$ t^{pq}-1 = \prod_{j=0}^{pq-1}(t-\omega^j),\quad t^p-1 = \prod_{r=0}^{p-1}(t-\omega^{qr}),\quad t^q-1 = \prod_{s=0}^{q-1}(t-\omega^{ps}) $$ and we see that $$ \phi(t) =\frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)} =\prod_{\substack{0<j<pq\\ p,q\nmid j}}(t-\omega^j), $$ ie has roots $\omega^j$ for $j$ not divisible by $p$ or $q$.

(1') An alternative proof uses that both $t^p-1$ and $t^q-1$ divide $t^{pq}-1$: eg, $(t^{pq}-1)/(t^q-1)=1+t^q+t^{2q}+\cdots+t^{(p-1)q}$.

The common divisor of $t^p-1$ and $t^q-1$ is $t^{\gcd(p,q)}-1=t-1$. This can be shown using $\gcd(t^p-1,t^q-1)=\gcd(t^{p-q}-1,t^q-1)$ where we assume $p>q$, and $\gcd(p,q)=\gcd(p-q,q)$, leading to Euler's algorithm.

Now, $(t^p-1)/(t-1)$ and $(t^q-1)/(t-1)$ have no common factor, and both divide $(t^{pq}-1)/(t-1)$, and so their product must also divide it: ie, $$ \frac{(t^{pq}-1)(t-1)}{(t^p-1)(t^q-1)} = \frac{\frac{t^{pq}-1}{t-1}}{\frac{t^p-1}{t-1}\cdot\frac{t^q-1}{t-1}} \quad\text{is a polynomial.} $$

(2) Next, to show that the coefficients are all in $\{0,1,-1\}$, let's start expanding as a formal power series $$ \frac{1-t^{pq}}{1-t^q}\cdot\frac{1}{1-t^p} =(1+t^q+\cdots+t^{(p-1)q})\cdot(1+t^p+t^{2p}+\cdots) =\sum_{j=0}^\infty\sum_{k=0}^{p-1} t^{jp+kq} =\sum_{n=0}^\infty a_nt^n. $$ In this, each power of $t$ appears at most once: if $jp+kq=j'p+k'q$ for some $j<j'$, $k>k'$, then $(k-k')q=(j'-j)p$ which is not possible as $0<k-k'<p$ and thus $k-k'$ cannot be divisible by $p$. Thus, $a_n\in\{0,1\}$. In fact, each power $t^n$ appears exactly once except from a finite number of $n$ which cannot be expressed as $n=jp+kq$, so actually $1/(1-t)-(1-t^{pq})/(1-t^p)(1-t^q)$ is a polynomial, although I will not use that result to complete my proof.

When we multiply the above power series by $1-t$, we get $$ \phi(t) = (1-t)\cdot\sum_{n=0}^\infty a_nt^n = \sum_{n=0}^\infty (a_n-a_{n-pq})t^n $$ where we $a_n=0$ for $n<0$. Since $a_n\in\{0,1\}$, the difference $a_n-a_{n-pq}\in\{0,1,-1\}$.

However, from (1) we know that $\phi(t)$ is a polynomial of degree $(p-1)(q-1)$, so for $n>(p-1)(q-1)$ we must have $a_n-a_{n-pq}=0$. The remaining coefficients of $\phi(t)$ are still in $\{0,1,-1\}$, which completes the proof.

(3) I thought the result could be generalised to cases where $m=\gcd(p,q)>1$, but $\omega^{mk}$ for $k=1,\ldots,pq/m-1$ are roots for both polynomials in the denominator, but only once in the numerator, so this would never work.