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Prove that $\forall a, b \in \mathbb{Z}^+. a \neq b. \exists \text{ infinite } n \in \mathbb{Z}^+, \gcd(a+n, b+n) = 1$

This question was asking to prove that for every two distinct positive numbers $a$ and $b$ there exist infinite positive numbers such that $\gcd(a +n, b+n) = 1$.

I tried proving it. This is my proof.

Proof:

Lemma 1: There exist infinitely many relative primes to a number which isn't 0. Let the number be $m$ and $m \neq 0$. Let $n$ range over all integers and $r$ be such that $0 < r < m$. So the set $S = \{mn + r\}$ consists of all such numbers.

Let $a$ and $b$ be the two numbers. Since they are different let us assume that a is the lesser one. So, $b = a + k$ and $k> 0$.

We know for every number, except 0, there exist infinitely many relatively prime numbers but none of $a, b, k$ are 0 over here by Lemma 1.

Let $s$ be any of those numbers and $s = a + n$.

So,

$$\gcd(a + n, k) = 1 \Longleftrightarrow \gcd(a+n, a+n +k) = 1 \Longleftrightarrow \gcd(a+n , b+n ) = 1$$

Thus, the proposition is proved.

Is my proof correct? Could you suggest some better way to prove this?

Bill Dubuque
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TheRandomGuy
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    The proof seems to work. However, be a bit clearer on exactly what criteria you want $s$ to fulfill when you pick it. It's supposed to be relatively prime to $k$, right? Then say that explicitly. – Arthur Mar 17 '16 at 09:24
  • @Arthur I am confused there. You see $s$ can be a complete set but I don't know how to show that. – TheRandomGuy Mar 17 '16 at 09:25
  • @Arthur Thanks. So I explicitly declare that $s$ is relatively prime to $k$. – TheRandomGuy Mar 17 '16 at 09:26
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    The latter part of your proof hinges on the fact that 1) $s > a$, so that you may write $s = a + n$, and 2) $\gcd(s, k) = 1$ so that you can do the last line of reasoning. You need to say those two things explicitly, and prove that there are infinitely many such $s$ (for instance, all primes greater than both $a$ and $k$ may be taken as such an $s$, and Euclid proved that there are infinitely many of those). – Arthur Mar 17 '16 at 09:28
  • @Arthur I have that fact that there are infinitely many $s$ prepared. I'll just add it in the question. – TheRandomGuy Mar 17 '16 at 09:29
  • If I were to prove this, I would pick a prime number $p > \max(a,b)$ and set $n = p-\max(a,b)$. – dtldarek Mar 17 '16 at 09:32
  • @Arthur I have added the lemma. – TheRandomGuy Mar 17 '16 at 09:32
  • @dtldarek Could you explain what you meant? I didn't get that. – TheRandomGuy Mar 17 '16 at 09:35
  • See here for a generalization to three gcd arguments. – Bill Dubuque Mar 10 '22 at 20:26

1 Answers1

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Your proof seems to work, but here is (in my opinion) a simpler one:

Let $p > \max(a,b)$ be any prime number strictly bigger than both $a$ and $b$, and then set $$n = p- \max(a,b).$$

Suppose that, $a > b$, then $n = p-a$ and $$\gcd(a+n, b+n) = \gcd(p, b+n) = 1$$ because $b+n < a+n = p$ and $p$ is a prime number.

Of course, there are infinitely many prime numbers bigger than $\max(a,b)$.

I hope this helps $\ddot\smile$

dtldarek
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    @Dhruv You were asking if there was a simpler way, so here it is. Yet, I would advise you to stick with your proof, because it is your own and with minor improvements it should be just fine. – dtldarek Mar 17 '16 at 09:45