Prove that $\gcd (a, b) = \gcd (a, b + xa)$ for any $x \in \mathbb{Z}$. 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 $u \mid t$. 
 A: 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.
