Prove that, there infinitely many terms have exactly $m$ distinct prime factors

Question

Let $X_n$ be an infinite arithmetic sequence with positive integers term. The first term is divisible by the common difference of successive members. Suppose, the term $x_i$ has exactly $m>1$ distinct prime factors, for some $i\in\Bbb N$.

Prove that, there infinitely many terms have exactly $m$ distinct prime factors.

My some thoughts.

Let $x_1$ has $n$ distinct prime factors and $x_2=x_1+d$ where $d\mid x_1\implies x_1=dk$. Hence $x_2=x_1+d=d(1+k)$. But, we can not say anything about the prime factors of $x_2$. Therefore, we don't know the number of prime factors of $x_2$. By induction we don't know the number of prime factors of $x_3$.

I tried to make the solution by induction, but as it seems I could not be successful.

Answer 1

We can be blunt.

Just prove there are are an infinite number of terms with the exact same prime factors (and thus the same number of factors)

All terms or multiples of $d$ and all multiples of $d$ that are equal to or greater than $x_1$ are terms. So if $x_i$ is a multiple of $d$ that is $\ge x_1$ with a unique factorization of $\prod p_i^{a_i}$ then any $\prod p_i^{b_i}; b_i \ge a_i$ will be a multiple of $d$ that is $\ge x_1$ so is also a term of the sequence.

And clearly there are an infinite number of them.

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To put it in the terms you started:

Let $x_{w} = d(k+w-1)$. Then $x_{w+(k+w-1)^j-(k+w-1)} = d(k+w-1)^j$ will have the exact same prime factors as $x_w$. And there are an infinite number of $j$s. The end.

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Note these are way overkill. But if overkill makes things easier than so be it. "Welcome to New York City! Our population is larger than 17!"

Answer 2

Clearly the AP is $a_{n+1} = d\cdot(k+n)$ for some $k,d \in \mathbb{N}$. Let $x_1 = d(k+r)$ for some $ r \in \mathbb{N}$, and let $x_t = d(k + ((k+r)^t - k))$ for any $t \in \mathbb{N}$. Then $x_t$ has the exact same prime factors as $x_1$. To show that this an infinite sequence, we only need to show that the terms are all distinct which is equivalent to $k + r = 1$. But as $k, r \ge 1$ we have $k + r \ge 2$, qed.