Show that either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$,but not both together.

Question

Let a complex number $\alpha,\alpha\neq1$,be a root of the equation $z^{p+q}-z^p-z^q+1=0$,where $p$ and $q$ are distinct primes.Show that either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$,but not both together.

Since $\alpha$ is a root of $z^{p+q}-z^p-z^q+1=0$.Therefore,$\alpha^{p+q}-\alpha^p-\alpha^q+1=0$ factorised to $(\alpha^p-1)(\alpha^q-1)=0$,but since $\alpha\neq1$

Therefore,either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$ or both equal to zero.Because we have studied that if $ab=0$,then either $a=0$ or $b=0$ or both equal to zero.But this is in contradiction to what is to be proved.My query is why cannot both be zero?

See http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 — lab bhattacharjee, Aug 15 '15 at 05:29
@labbhattacharjee,I cant co-relate the two questions,the link you gave and my question. — mathspuzzle, Aug 15 '15 at 05:39
$$\sum_{r=0}^{n-1}\alpha^r=\dfrac{\alpha^n-1}{\alpha-1}$$ right? — lab bhattacharjee, Aug 15 '15 at 05:58

score 4 · Accepted Answer · answered Aug 15 '15 at 05:58

$z= \cos \theta+ i\sin \theta\to z^p = \cos p\theta+i\sin p\theta=1, z^q=1\to \cos q\theta + i\sin q\theta=1\to p\theta = 2m\pi, q\theta = 2n\pi\to \dfrac{m}{p}=\dfrac{n}{q}\to m = kp, n = kq\to \theta = 2k\pi\to z = 1$, contradiction since $z \neq 1$. Thus they cannot be both $0$.

score 2 · Answer 2 · answered Aug 15 '15 at 08:02

Without even knowing the form of the roots: \begin{align*}z^{p+q}-z^p-z^q+1&=(z^p-1)(z^q-1)\\&=(z-1^2)(1+z+z^2+.....+z^{p-1})(1+z+z^2+.....+z^{q-1})\end{align*} Hence if $\alpha\neq 1$, $\;1+\alpha+\alpha^2+.....+\alpha^{p-1}=0\;$ or $\;1+\alpha+\alpha^2+.....+\alpha^{q-1}$.

Not both: If both, then we also have both $\;\alpha^p-1=0\;$ and $\;\alpha^q-1=0$. However, it is a classic fact that $$\gcd(z^p-1,z^q-1)=z^{\gcd(p,q)}-1=z-1,$$ so that, by Bézout's identity, there exist polynomials $u(z), v(z)$ such that $$1=u(z)(z^p-1)+v(z)(z^q-1),\enspace\text{whence}\quad 1=u(\alpha)(\alpha^p-1)+v(\alpha)(\alpha^q-1)=0 $$ which is absurd.

score 1 · Answer 3 · answered Aug 15 '15 at 08:18

Berenard's answer with a differtn perspective: Let $I=\{\,n\in\mathbb Z\mid \alpha^n=1\,\}$. Trivially, $0\in I$. If $n\in I$ then also $-n\in I$. And if $n,m\in I$ then $\alpha^{n+m}=\alpha^n\alpha^m=1$ so that also $n+m\in I$. We conclude that $I$ is a subgroup of $\mathbb Z$, hence of the form $I=d\mathbb Z$ with $d\in\mathbb N_0$ Now if $\alpha$ is a root of $z^{p+q}-z^p-z^q+1$ then you already shows that $p\in I$ or $q\in I$. We are also given that $1\notin I$, i.e., $d\ne 1$. Then $p\in I$ implies $d\mid p$ and so $d=p$. Similarly $q\in I$ implies $d=q$. As $p\ne q$, we cannot have both $p\in I$ and $q\in I$.

score 0 · Answer 4 · edited Apr 13 '17 at 12:20

If $a$ is a root of $1+z+z^2+\cdots+z^{m-1}=0\ \ \ \ (1),a\ne1$

and is a root of $\dfrac{z^m-1}{z-1}=0$

If $a$ is a root of $1+z+z^2+\cdots+z^{n-1}=0\ \ \ \ (2),a$ is a root of is a root of $\dfrac{z^n-1}{z-1}=0$

Now Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ says $(a^m-1,a^n-1)=a^{(m,n)}-1$

$$\implies\left(\dfrac{a^m-1}{a-1},\dfrac{a^n-1}{a-1}\right)=\dfrac{a^{(m,n)}-1}{a-1}$$

If $(m,n)=1,$

$$\left(\dfrac{a^m-1}{a-1},\dfrac{a^n-1}{a-1}\right)=1$$

$\implies(1),(2)$ can not have any root in common

Show that either $1+\alpha+\alpha^2+.....+\alpha^{p-1}=0$ or $1+\alpha+\alpha^2+.....+\alpha^{q-1}=0$,but not both together.

4 Answers4