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 $x^2 - 1 \mid x^{2n} -1$.