# A question about proving distinct Fermat numbers are relatively prime

I am reading Elementary Number Theory with Programming, and bump into the proof of "The Fermat numbers $$F_n$$ and $$F_m$$ where m ≠ n are relatively prime.":

Theorem: The Fermat numbers $$F_n$$ and $$F_m$$ where m ≠ n are relatively prime.

Proof: Let n > m and let d = gcd($$F_n$$, $$F_m$$). Since $$F_n$$ and $$F_m$$ are odd, it follows that d is odd. We will show that d = 1. Let x = $$2^{2^{m}}$$ and let a = $$2^{n−m}$$. Then $$F_n$$ −2 = $$2^{2^{n}}$$ −1 = $$x^a$$ −1. Since a is even, x = −1 is a root of the equation $$x^a$$ − 1 = 0, implying that x + 1 | $$x^a$$ − 1. Then $$F_m$$| $$F_n$$− 2. Since d | $$F_m$$, it follows that d | $$F_n$$ − 2. Now, since, in addition, d | $$F_n$$, we have d | 2, implying that d = 1 since d is odd, and the proof is over.

I can understand all except the following words:

Since a is even, x = −1 is a root of the equation $$x^a$$ − 1 = 0, implying that x + 1 | $$x^a$$ − 1.

How is "x + 1 | $$x^a$$ − 1" concluded?

Hint. $x^2-1 = (x-1)(x+1)$ and $x^2 - 1 \mid x^{2n} -1$.
• I still can't figure out why "x = −1 is a root of the equation $x^a$ − 1 = 0" can deduce "x + 1 | $x^a$ - 1". What is the relationship between these 2 statements. Jan 26, 2018 at 9:17
• Well, if $b$ is a root of a polynomial $P$, then $X-b$ divides $P$. Indeed, do the Euclidian division, $P = Q(X-b) + R$, with $d(R) < d(X-b) = 1$, that is $d(R) = 0$. Since $P(b) = 0$, $R = 0$. Jan 26, 2018 at 9:25