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  • $\begingroup$ Thanks this one is nice. And thanks to the other post I got to know about the nice theorem by Evertse, Schmidt and Schlikewei. +1. $\endgroup$ Commented Jun 22, 2014 at 16:11
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    $\begingroup$ I made it simple by ignoring the character of d. But if d has all the primes below p as factors, then one of the first (two or three) terms has a prime factor about as big as p or bigger anyway, so a positive progression of length n should have a term divisible by a prime q at least as big as p, which is the largest prime less than n. $\endgroup$ Commented Jun 22, 2014 at 16:17
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    $\begingroup$ For any arithmetic progression $a,a+d,\dots,a+(m-1)d$ supported on primes at most $N$, you prove that its length $m$ is less than $p$, where $p>N$ is the smallest prime not dividing $d$. How do we get a uniform bound (depending only on $N$) from here? I could not follow your comment above. $\endgroup$ Commented Jun 22, 2014 at 23:04
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    $\begingroup$ @GHfromMO, noted. I hope the edit to follow satisfies you. $\endgroup$ Commented Jun 23, 2014 at 5:40