Animated Solution for Mathematics - Complex Numbers: Let a complex number $\alpha ,\alpha \ne 1$, be a root of the equation $z^{p+q}-z^p-z^q+1=0$, where $p,q$ are distinct primes. Show that either $1+\alpha +{\alpha }^2+\cdots +{\alpha }^{p-1}=0$ or $1+\alpha +{\alpha }^2+\cdots +{\alpha }^{q-1}=0$, but not both together.

Welcome, future engineer! Today we are going to dissect a problem that sits at the beautiful intersection of algebra and number theory. It is a problem that might look like a daunting beast at first, but once you peel back the layers, you will find a core of pure, elegant logic.

Let us begin by looking at this fascinating complex equation:

Here, $p$ and $q$ are distinct prime numbers, and we are given a complex root $\alpha$ which is definitely not equal to $1$. Our goal is to show that $\alpha$ satisfies one of two geometric series sums, but absolutely cannot satisfy both at the same time.

Look at the structure of the polynomial. We can simplify it by grouping the terms:

This reveals a common factor of $(z^q-1)$. We can write the entire equation in a beautifully factored form:

Since the product of these two factors is zero, mathematics tells us that at least one of the individual factors must be zero. Therefore, any root $\alpha$ of our original equation must satisfy either ${\alpha }^p=1$ or ${\alpha }^q=1$.

Any complex number satisfying $z^n=1$ is called an $n$-th root of unity, and all such roots lie perfectly on a circle of radius $1$, centered at the origin. The point $1$ is a common root for both sets, but our problem explicitly states that $\alpha eq 1$.

If $\alpha$ is a root of $z^p=1$ and $\alpha eq 1$, the sum of the terms $1+\alpha +{\alpha }^2+\cdots +{\alpha }^{p-1}$ is given by:

Since ${\alpha }^p=1$, the numerator becomes zero, making the entire sum equal to zero. By applying the exact same logic to the second factor, if $\alpha$ satisfies ${\alpha }^q=1$, then the sum of the geometric series $1+\alpha +{\alpha }^2+\cdots +{\alpha }^{q-1}$ must also be equal to zero.

To prove that both sums cannot be zero simultaneously, we use a proof by contradiction. Suppose that both sums are zero, meaning ${\alpha }^p=1$ and ${\alpha }^q=1$.

Because $p$ and $q$ are distinct prime numbers, their greatest common divisor is $\gcd (p,q)=1$. By Bézout's Identity, there exist integers $a$ and $b$ such that:

Now, let us raise $\alpha$ to the power of $1$:

Substituting the assumed values ${\alpha }^p=1$ and ${\alpha }^q=1$, we get:

This is a direct contradiction to our initial constraint that $\alpha eq 1$. Therefore, our assumption that $\alpha$ could satisfy both equations simultaneously must be false. We have successfully proven that $\alpha$ satisfies exactly one of the two geometric sums.