Multiples of Factorial Plus One are Coprime

Theorem

Let $a \in \N$ be a natural number.

Let $\sequence {x_1, x_2, \dotsc, x_n}$ be a sequence of natural numbers such that:

$\forall i: 1 \le i \le n: x_i = 1 + i \times a!$

where $a!$ denotes the factorial of $a$.

Let $a \ge n - 1$.

Then $\set {x_i}$ are pairwise coprime.

Proof

Let $p$ be a prime number.

Suppose that $p \divides x_i$.

Then, by definition of $x_i$:

$p \divides 1 + i \times a!$

Thus, by Consecutive Integers are Coprime:

$p \nmid i \times a!$

Therefore:

$p \nmid a!$

Aiming for a contradiction, suppose $p \divides x_j$, where $j \ne i$.

Then $p \divides 1 + j \times a!$.

By Common Divisor Divides Difference:

$p \divides \paren {i - j} \times a!$

By Euclid's Lemma, along with $p \nmid a!$ above:

$p \divides i - j$

But as $0 < \size {i - j} < n$:

$p \divides \paren {n - 1}!$

Additionally, since $a \ge n - 1$:

$\paren {n - 1}! \divides a!$

By Divisor Relation is Transitive:

$p \divides a!$

However, $p \nmid a!$ from above, a contradiction.

Thus, by Proof by Contradiction, there is no such $x_j$.

Hence the result.

$\blacksquare$