$(m,n)=1$, $a^m=1$, $b^n=1$, can we show $a+b+1\neq 0$?

Question

Let $m,n$ be two natural numbers, $a,b$ are two complex numbers, and the greatest common divisor of $m$ and $n$ is $1$: $(m,n)=1$, $a^m=1$, $b^n=1$, can we show $a+b+1\neq 0$?

For small $m,n$, I calculate so it is right. But I do not know whether it is true, and find a proof.

Answer 1

Suppose $a$ and $b$ are roots of unity with $a+b=-1$. Then $\Im(a)=-\Im(b)$. But now $|a|=|b|=1$, so that $\Re(a)=\pm\Re(b)$. As $a+b=-1$, we must have $\Re(a)=\Re(b)=-1/2$. Writing $a=\cos(\theta)+i\sin(\theta)$, we see that $\theta=2\pi/3$ or $\theta=4\pi/3$. In particular, $a$ and $b$ must be $3$rd roots of unity.

We may conclude that if $a$ and $b$ are roots of unity, whose orders are coprime, then $a+b\neq -1$.

Answer 2

Since $a^m$ and $b^n$ are both equal to 1, then we can infer that $|a|=|b|=1$. So instead of $a$ and $b$, we'll call them:

$a=e^{i \theta_1}, b=e^{i \theta_2}$

Let's assume $a+b = -1$, and let's call $|\theta_1 - \theta_2| = \alpha$. The magnitude of $|a+b|=1$, so $\alpha$ must be 120 degrees or $2\pi / 3$ radians. Because: (I'm treating the numbers as vectors)

$1=\sqrt{1^2 + 1^2 + 2 \times 1 \times 1 \times \cos(\alpha)} => \alpha=\dfrac{2\pi}{3}$

Let's assume $\theta_2 > \theta_1$, so: $\theta_2 = \theta_1 + \dfrac{2\pi}{3}$

Now we have to solve for the following equations:

$\cos(\theta_1) + \cos(\theta_1 + 2\pi/3) = -1$

$\sin(\theta_1) + \sin(\theta_1 + 2\pi/3) = 0$

The only solution is when $\theta_1 = \dfrac{2\pi}{3}$.(Assuming $\theta \in [0,2\pi]$)

So we conclude that $a+b=-1$ iff $a$ and $b$ are the following(respectively):

$e^{2\pi/3 \times i}, e^{4\pi/3 \times i}$, or just $\dfrac{-1+\sqrt{3}i}{2}$ and $\dfrac{-1-\sqrt{3}i}{2}$.

But these two numbers are both the answer to the equation $x^2+x+1$, which means $x^3-1$ is zero. So $(m, n)=3$ which contradicts the initial conditions of the question. So $a+b \neq -1$