Proof Verification, Theorem due to G Polya Problem:[G. Polya] Let $a$ and $b$ be two relatively prime positive integers, and consider the arithmetic progression $a, a + b, a + 2b, a + 3b, . . . .$ Prove that there are infinitely many terms in the arithmetic progression that have the same prime divisors.
My Attempt: Let the general term of this sequence be $x\equiv a\pmod b.$ Then for all $i\geq 1,$ $x_i=ax^{i\phi(b)}$ is a member of this sequence. If $x=p_1^{k_1}p_2^{k_2}p_3^{k_3}...p_r^{k_r}$ then $x_i$ has the same prime divisors for all $i\geq 1.$
I would like to know whether this proof is correct or not because the solution in the textbook I am using is different and since I am a beginner in Number Theory there might be subtle details that I may have overlooked.
 A: I think you may have mis-read the theorem's conclusion.  I believe it is not saying that there is a single prime which divides infinitely many terms in the progression, I believe it is making the much stronger statement that there is a set of primes, and an infinite number of terms of the progression such that each term is divisibly by the primes in the set -and no other primes-.  For example, there may be the prime set { 2, 3, 5 }, and infinitely many terms of the progression are divisible by 2, 3, and/or 5, but not by 7, 11, ... .
of course I could be wrong...
A: I suspect PMar is correct about the interpretation of the question: we should show there exists a set of primes $S$ such that infinitely many terms in the arithmetic progression have exactly $S$ as their set of prime divisors. In fact, if $a$ is relatively prime to $b$, then you should be able to show the set of powers $\{a,a^2,a^3,\cdots\}$ has infinite intersection with $\{a,a+b,a+2b,\cdots\}$. (Hint: the sequence of powers is repeating mod $b$ and must return back to $a$ since $a$ is a unit mod $b$.)
A: The question is asking us to prove there exist infinitely many $(i,j)$ so that prime $p|a+ib \iff p|a+jb$
Pick a prime $p \equiv 1\ \pmod b$. It's existence is guaranteed due to Dirichlet as $\gcd(1,b)=1$.
Then $ap, ap^2, ap^3,...$ all have the same prime divisors, namely prime divisors of $a$ and $p$.
