Emirhan wrote: An $a_n$ sequence is a arithmetic sequence with all terms be positive integers. Let $p_n$ is greatest prime divisor of $a_n$. Prove that $(\frac{a_n}{p_n})$ sequence is infinity.

I assume the arithmetic sequence is supposed to be nonconstant?

Yes. $a_n$ is a non-constant sequence. Thanks for letting me know

Actually arithmetic sequence condition is not necessary. The only condition that we need is, sequence $\big( a_n\big )$ of distinct natural numbers for which $a_{n+1}-a_n, n=1,2,\dots$ is bounded. For the solution, just consider a natural number $x$ and primes $\big( q_i\big )$ large enough which satisfies $q_i^2 | x+i, i=0,1,\dots ,d$ where $d$ is bigger than the upper bound of $a_{n+1}-a_n.$ Now we can see that there must be a natural number $n$ which $a_n$ is equal to one of $x+i$. Since $q_i^2 | a_n$, we have $\frac{a_n}{p_n} \geq q_i$ $\square$

We are going to prove there is a prime $p$ which is greater than a positive integer $g$ such that there are infinitely many positive integer $n$ which concludes $a_n = gp^k$. Let $a_n = a_0 + nd$. Let $g = (a_0, d)$. So for $(x, y) = 1$, $a_n = g(x + yn)$. Let's define $b_n = x + yn$. By dirichlet theorem, we can choose some prime $p$ such that for any $t$, $b_t = p$ and $p > g$, $p > y$. Now the cay claim comes: Claim: İf there are positive integers $i$, $l$ such that $b_i = p^l$ , there is positive integer $j > i$ which provides $b_j = p^{l + \phi(y)}$. Proof: We want to show for some $j$, $b_j - b_i = p^l(p^{\phi(y)}-1)$. That means we are trying to find some positive integer $j$ such that $y(j-i) = p^l(p^{\phi(y)}-1)$ . Actually $j = i + \frac{p^l(p^{\phi(y)}-1)}{y}$ is really obvious since $y | p^{\phi(y)}-1$ because $p > y$. By the claim, some terms of the $a_n$ are: $$gp, gp^{\phi(y) + 1}, gp^{2\phi(y) + 1}, gp^{3\phi(y) + 1} ...$$and we know that for all these terms, the greatest prime divisor is $p$ since $p > g$. So if we divide these all terms to $p$, the result is clear. $\blacksquare$