TITLE: Arbitrarily long arithmetic progressions
QUESTION [10 upvotes]: Are there arbitrarily long arithmetic progressions in which all the
  prime factors of all the terms are at most $N$, for some $N$? Assume
  all the terms are positive and the sequence of terms is increasing.

I have proved that no such infinite sequence exists. Note the $N$ may vary from AP to AP.
For infinite sequences let $\{a+nd\}_{n\ge 0}$ be the AP. $\text{gcd}(a,d)=s$ then $a+nd=s(x+ny)$ for some $x,y$ with $\text{gcd}(x,y)=1$ then by Dirichlet's theorem $\{x+ny\}_{n\ge 0}$ has infinitely many primes. Thus we have the prime factors of the sequence is unbounded and hence done. But I was thinking about this claim but nothing came in mind. Could someone help? Thanks a lot.

REPLY [3 votes]: This is a supplement to the elementary proofs given by The Masked Avenger and Lucia.
Theorem. For any $k$ there is $m$ such that no arithmetic progression of length $m$ is supported on $k$ distinct primes.
Proof. We prove by induction on $k$. For $k=0$ we can clearly take $m=2$. So let $k\geq 1$ and assume the statement for $k-1$ in place of $k$. That is, there is $m$ such that no arithmetic progression of length $m$ is supported on $k-1$ distinct primes. Consider an arithmetic progression of length $m'$ supported on the $k$ distinct primes $p_1,\dots,p_k$. Without loss of generality, the terms are coprime to the difference $d$. Observe that in any consecutive block of length $m$ in the progression, each prime $p_i$ occurs as a divisor by the induction hypothesis. Hence for $m'\geq m$ we get that $p_i\nmid d$, and for $m'\geq 2m$ we get that $p_i\mid kd$ for some $0<k<2m$. Combining the two statements, we infer that for $m'\geq 2m$ each $p_i$ is less than $2m$. Using Lucia's elementary argument, it follows that $m'<\prod_{p<2m}p<2^{4m}$. That is, for $m':=2^{4m}$ we get a contradiction, proving that no arithmetic progression of length $m'$ is supported on $k$ distinct primes.
Remarks (updated). The modern results on $S$-unit equations mentioned by Felipe Voloch yield an exponential bound on $m$. Lucia's argument combined with Iwaniec's bound on Jacobsthal's function yields directly $m\ll k^2(\log k)^2$, see Demonstratio Math. 11 (1978), 225–231. Weaker versions of this bound admit simple elementary proofs.