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This question is a possible duplicate, but I don't understand the answers.

Let $a, b, c \in \mathbb{Z}$. Prove $\gcd (a, b) = \gcd (a, b + xa)$ for any $x \in \mathbb{Z}$.

This is my definition of the $\gcd$: The greatest common divisor of integers $r$ and $s$ is an integer $t$ such that 1. $t \mid r$ and $t \mid s$ 2. For any integer $u$, if $u \mid r$ and $u \mid s$, then $u \mid t$.

I have a few questions about proving this. To do it, do I suppose that $t = \gcd (a, b + xa)$ and then show that $t = \gcd (a, b)$. Will I be done then? Or do I need to do that and also suppose $t = \gcd(a, b)$ and then show that $t = \gcd(a, b + xa)$? (I don't think that is necessary, but I want to make sure. I think I am getting confused with proving $A \iff B$)?

I am also confused with the wording "for any $z \in \mathbb{Z}$." Does for any mean for all $z$? Or does it mean that $z$ is fixed?

Here's my attempt: Suppose $t = \gcd(a, b + xa)$. Then $t \mid a$ and $t \mid (b + xa)$ and so $a = tj$ and $(b + xa) = tk$ for some integers $j$ and $k$. Thus $b = tk - xa = tk - tja = t(k - ja)$. Thus $t \mid b$. Suppose $u$ is an integer such that $u \mid a$ and $u \mid b$. Then $a = u \alpha$ and $b = u \beta$ for some $\alpha, beta \in \mathbb{Z}$. Since $t = \gcd(a, b +xa)$, we have that $t = am + (b + xa)n$ for some integers $m$ and $n$. Then $t = am + bn + xan = u \alpha m + u \beta n + u x \alpha n = u(\alpha m + \beta n + x \alpha n)$. Thus $u \mid t$ and so $t = \gcd(a, b)$.

I said earlier that I am not sure if I need to conversely suppose that $t = \gcd (a, b)$ and show that $t = \gcd (a, b + xa)$. I will go ahead and work on that proof so in case it is necessary, I can get some feed back. (Or an explanation on why it is redundant.)

Suppose $t = \gcd(a, b)$. Then $t \mid a$ and $t \mid b$ so $t \mid (yb + xa)$ for all $x, y \in \mathbb{Z}$. Taking $y = 1$, we have that $t \mid (b + xa)$. Suppose $u \in \mathbb{Z}$ such that $u \mid a$ and $u \mid (b+xa)$. Then $a = u \alpha$ and $b + xa = u \beta$. I'm stuck on showing that $u \mid t$. I know I can write $t = am + bn$ for $m, n \in \mathbb{Z}$, but I can't figure out how to show that $u \mid t$.

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marked as duplicate by Jyrki Lahtonen Jun 30 '16 at 12:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Did you mean "for any $x\in\mathbb Z$"? $\endgroup$ – NoetherianCheese Mar 8 '16 at 2:48
  • $\begingroup$ @Cheese Yes. I will edit. $\endgroup$ – user100000000000000 Mar 8 '16 at 2:50
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    $\begingroup$ There's a logical subtlety involved here: the definition of the gcd that you give doesn't directly imply uniqueness. Indeed, over the integers, the gcd is not unique (if $x$ is a gcd, then so is $-x$) and over more general structures (e.g. a field) there might be even more gcds. Yet the notation $x=gcd(a,b)$ is often (ab-)used. That's why in theory you'd have to show both directions of the implication. In practice, though, you don't, because the gcd is unique up to the sign. $\endgroup$ – Fryie Mar 8 '16 at 3:03
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    $\begingroup$ @Fryie makes an important point. In an integral domain (a set with an addition and multiplication which work a lot like the integers) GCDs are only determined up to associates (they can be off by an invertible element). In the context of the integers, your definition allows a positive and negative GCD for each pair. However, it is a standard convention to pick the positive one. For example: Your definition allows both $\pm 4$ to be the GCDs of $12$ and $20$, but our usual conventions say that $4$ is the GCD. Similarly when working with polynomials we choose the GCD to be a monic. :) $\endgroup$ – Bill Cook Mar 8 '16 at 3:09
  • $\begingroup$ Even without that convention, the first part of the proof works fine (the second is not needed) since it shows that if $t$ is a gcd of $(a,b+xa)$, it is also one of $(a,b)$. To show that the two sets of gcds are equal, it is sufficient to know that $t$ is a gcd iff $-t$ is (and there is no other gcd), i.e. one gcd uniquely determines all the others. $\endgroup$ – Fryie Mar 8 '16 at 3:24
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First, yes, "for any" and "for all" are essentially interchangable.

Next, you have a typo in one of your lines $b=tk−xa=tk−tja=t(k−ja)$ should be $b=tk-xa=tk-xtj=t(k-xj)$ (but you have the right idea).

Your proof looks fine. If you assume the GCD of $a$ and $b+xa$ is $t$, then go on to show that the GCD of $a$ and $b$ is $t$, you've shown they have the same GCD and you're done.

I might approach this proof a little differently myself. To show that two pairs of numbers have the same GCD, just show that they have the same set of common divisors (thus they'll share the same GCD). I believe this looks a little cleaner:

Suppose that $d$ divides both $a$ and $b$. Then there are integers $k,\ell$ such that $dk=a$ and $d\ell=b$. Thus $b+xa=d\ell+xdk=d(\ell+xk)$ so that $d$ divides $b+xa$.

Conversely, suppose that $d$ divides both $a$ and $b+xa$. Then there are integers $m,n$ such that $dm=a$ and $dn=b+xa$. Thus $b=dn-xa=dn-xdm=d(n-xm)$ so that $d$ divides $b$.

Therefore, if $d$ is a common divisor of $a$ and $b$, it is a common divisor of $a$ and $b+xa$. Likewise, if $d$ is a common divisor of $a$ and $b+xa$, then it is a common divisor of $a$ and $b$.

Since $a,b$ and $a,b+xa$ share the same common divisors, they must have the same greatest common divisor.

EDIT: To address a comment above. Your definition of a GCD does not uniquely determine an integer.

For example: Both $\pm 4$ are GCDs of $12$ and $20$. However, when working with integers, we usually pick the positive one to be THE GCD. But technically, your definition allows for both answers.

If you go with this "understood" assumption, your proof is ok (up to correcting a typo or two).

My proof works over any integral domain (after changing "integers" to "ring elements") and shows that any GCD of $a,b$ is also a GCD of $a,b+xa$ for any ring element $x$ (and conversely). In fact, if you pick out a specific GCD (like a positive integer or monic polynomial) the proof still works.

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    $\begingroup$ +1: This answer gives good insight and even more than was asked, and it circumvents all the problem with uniqueness and such. $\endgroup$ – Fryie Mar 8 '16 at 3:29
  • $\begingroup$ @Bill Cook Thank you so much! That helped me so much! $\endgroup$ – user100000000000000 Mar 8 '16 at 3:46
  • $\begingroup$ Glad I could help! :) $\endgroup$ – Bill Cook Mar 8 '16 at 4:02

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