Prove for some $n\in\mathbb{N}$ there are infinitely many primes $p$ , S.T the numbers $p-1,p+1,p+2$ have $n$ different prime factors .

Question

Prove for some $n\in\mathbb{N}$ there are infinitely many primes $p$ , S.T the numbers $p-1,p+1,p+2$ have $n$ different prime factors .

Attempt:

by the fundamental theorem of arithmetic's

we know that $$p-1=p_1^{a_1}...p_k^{a_n}$$ also $$p+1=q_1^{a_1}...q_k^{a_n}$$ $$p+2=w_1^{a_1}...w_k^{a_n}$$

Noticing that all the factorizations consists of $n$ different prime factors.

1. I tried proving a stronger case with adding $p$ and its factorization to achieve 4 consecutive numbers. However that led me no where.
2. I tried to solve it with the Chinese remainder theorem using the abstract idea that there is only one unique $x_0$ solution. Not sure how to continue with this idea.

Answer 1

As pointed out in the comments: If the question is asking for exactly $n$ factors, then the claim is false for every $n$, for the following simple reason:

If $p \neq 2,$ then $p \equiv 1 \pmod{2},$ thus $p-1$ and $p+1$ are both divisible by $2$, thus can not have different prime factorizations.

However, the statement is true for at least $n$ different prime factors:

Let $p_1,...,p_n, q_1,...,q_n, r_1,...,r_n$ be different odd prime numbers.

Let $P = p_1p_2 \cdots p_n, Q =q_1q_2\cdots q_n$ and $ R= r_1r_2\cdots r_n.$

(1) By the Chinese Remainder Theorem, \begin{align*} x-1 &\equiv 0 &\pmod{P} \\ x+1 &\equiv 0 &\pmod{Q}\\ x+2 &\equiv 0 &\pmod{R} \\ \end{align*} has a solution.

(2) Let $x\in \mathbb{N}$ be a solution of the congruence equations. Note that $x_k = x+ k \cdot PQR $ is also a solution for all $k\in \mathbb{N}$.

(3) Now $(x,PQR) =1$, so by Dirichlet's theorem of primes in arithemetic progression you get the result.