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I get some troubles with this problem.

Initial Problem:

Given the infinite sequence $a+tb$ with $t\in\{0,1,2,\dots\}$ prove there exists an infinite subsequence such that all pairs of elements in the subsequence are relative primes.

I think the problem has an error on $a$ and $b$ and I will change that part.

Modified Problem:

Given the infinite sequence $a+tb$ with $t\in\{0,1,2,\dots\}$ and $\gcd(a,b)=1$, prove there exists an infinite subsequence such that all pairs of elements in the subsequence are relative primes.

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    are $a$ and $b$ coprime? – 2017-01-21
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    Well if $t=0$ and $a>1$ then this is false. What have you tried yourself? And where did you come across this problem? What kind of tools do you know/have? If $t>0$ and $a$, $b$ and $t$ are coprime then this is a trivial corollary of [Dirichlet's theorem on arithmetic progressions](https://en.wikipedia.org/wiki/Dirichlet's_theorem_on_arithmetic_progressions) – 2017-01-21
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    @Servaes I need a subsequence of the sequence $a+tb$, then $t=0$ is not in your subsequence. – 2017-01-21
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    @JorgeFernándezHidalgo Not necessarily, maybe and maybe not. I think is for all $a$ and $b$ becouse the exercise doesn't say anything about it. – 2017-01-21
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    @JoseA132 If $a=2, b=4$ then every $a+tb$ is even.... – 2017-01-21
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    @JoseA132 I do not understand your comment. If $t=0$ then your sequence is $a,a,a,\ldots$, so *no* two elements of the sequence are coprime if $a>1$. – 2017-01-21
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    @Servaes If $a$ and $b$ shouldn't be integers then they shouldn't be coprimes. $t$ could be the value you want, so you can always peek a value of $t$ coprime with $a$. – 2017-01-21

2 Answers 2

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If $a$ and $b$ are not coprime, this is false, because $\gcd(a,b)$ divides every term in the sequence.

If they are, a deep thorem by Dirichlet says that in the sequence $a+tb$ there are infinitely many primes. These primes can make the subsequence that you are looking for.

If you are not allowed to use this theorem, you can also show it this way.

Define the subsequence recursively. The first terms will be $a_1=a$, $a_2=a+b$.

Suppose that you have $n$ terms of the sequence that are pairwise coprime. Let $p_1,\ldots,p_s$ be the primes that divide any of these $n$ terms. Then define the $n+1$-th term like this: $$a_{n+1}=a+p_1p_2\cdots p_sb$$

Then, for $k=2,\ldots,n$, let $p$ be a prime common factor of $a_k$ and $a_{n+1}$. Since $p$ divides $a_k$, then $p=p_j$ for some $j$. So $p_j$ divides $a$. Then $p_j$ is a prime common factor of $a_1$ and $a_k$, a contradiction. This proves that the new term is coprime with the old ones.

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    Of course, but I think such a statement require far less theory than Dirichlet theorem on arithmetic progressions... – 2017-01-21
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    Well, OP said nothing about elementary proofs. Perhaps this works for him/her. – 2017-01-21
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You can just build the sequence recursively. Suppose that $a+t_1b>a+t_2b>\dots > a+t_nb$ have already been defined. You just need to find an $a+t_{n+1}b$ that is coprime to the previous ones.

This can always be done. we need to find an integer $x$ that is congruent to $a\bmod b$ and is coprime to a certain positive $M$ (here $M$ is the product of the previous terms).

This is always possible by the chinese remainder theorem. Let $p_1^{a_1}p_2^{a_2},\dots p_k^{a_k}=b$ and let $p_{k+1},\dots p_s$ be the primes that divide $M$ but not $b$.

The congruence system:

$x\equiv a \bmod p_1^{a_1}$

$x\equiv a \bmod p_2^{a_2}$

$\dots$

$x\equiv a \bmod p_k^{a_k}$

$x\equiv 1 \bmod p_{k+1}$

$\dots$

$x\equiv 1\bmod p_s$

Clearly gives us integers that are congruent to $a\bmod b$ and are coprime to $M$(because $a$ is coprime to $b$, so none of the first $k$ congruences implies divisibility by $p_i$). This allows us to continue building the sequence at each step.