Suppose $a$ and $b$ are relatively prime and $x \in \mathbb{Z}$. Show that if $a|bx$, them $a|x$ The hint for this question is "Use existence of $s,t$ such that $sa + tb = 1$"
So, the relevant definitions and corollaries that I have collected for this problem are:

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*Def of divisibility - Since $a|bx$ that means that $(bx) = qa$ with $q \in \mathbb{Z}$

*Def of relatively prime - GCD($a,b$) = $1$

*Corollary to (2.) - $\exists$ $s,t$ such that $sa + tb = 1$
Now, conceptually I believe I understand the logic for this statement: We know that $a$ divides the product $bx$ but we also know that $a$ and $b$ are relatively prime and hence $a$ cannot divide $b$ since the greatest divisor between them is $1$. This only leaves the factor of $x$ for which $a$ can divide in order to divide the product $bx$.
I'm having a bit of difficulty putting this into "math terms":

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*I started by writing out the relevant definitions above and seeing if there was any way I could combine them to show that $a$ divides $x$ or doesn't divide $b$. I started by writing out the def of relatively prime integers $sa + tb = 1$ I then multiplied by $x$ in order to introduce an $x$ into the equations then I subtracted the $tbx$ to get: $sa = x - tbx = x(1 - tb)$ The $x$'s just cancel and that brings us in a round-about way back to a single manipulation of the original equation $sa = 1 - tb$ which I don't think really says anything valuable or makes any headway to our goal.


*Next, I tried starting from the equation $sax + tbx = x$ again but this time substituting the def of divisibility above to get $sax + t(qa) = x$. From here, I pulled out the $a$ to get $a(sx + tq) = x$ which does look promising because if we notice that $(sx + tq) = k \in \mathbb{Z}$ then we can see that $ak = x$ is in the form for the definition of divisibility for $x$. So can we say that this proves that $a$ divides $x$? Furthermore, is that sufficient, if I state (in plain english) that $a$ cannot divide $b$ because of the reasons stated above? Or do I need to show that in "math terms" as well?
 A: Your second solution is great, and you can state that $a$ cannot divide $b$ because of the reasons above.
One tiny thing I would suggest you add, depending on how harsh the graders might be (if this were graded), would be to state why $sx$ and $tq$ are each integers, and why $k$ is therefore an integer.
A: We first observe that $\gcd(a, b) = 1$ implies $a \not \mid b$ unless in fact $a = 1$; otherwise we would have $\gcd(a, b) = a \ne 1$.
Given that
$\gcd(a, b) = 1, \tag 1$
it follows from Bexout's identity that there exist
$s, t \in \Bbb Z \tag 2$
with
$as + bt = 1; \tag 3$
we multiply this by $x$:
$asx + btx = x; \tag 4$
we further observe that
$a \mid bt \Longrightarrow \exists c \in \Bbb Z, \; bt = ac; \tag 5$
thus (4) may be transformed into
$asx + acx = x, \tag 6$
from which it immediately follows that
$a(sx + cx) = x \Longrightarrow a \mid x. \tag 7$
QED.
I agree with our colleague I am a person:  your second solution is the way to go.  In fact, it is more-or-less the "standard proof" of this "theorem".
