Prove that the Diophantine equation $ax+by+cz=e$ has a solution if and only if $(a,b,c)\mid e$. I have an intuitive idea about how this is going to work, but I don't know if I'm writing it properly using proper math language and theorems. I am most uncomfortable with the second half of the proof since it feels really weakly stated.

Let $(a,b,c)=d$. Therefore, there exists some $x,y,z\in\mathbb{Z}$ such that $ax+by+cz=d$. Additionally, $(\frac{a}{d},\frac{b}{d},\frac{c}{d})=1$, and there exists the same $x,y,z$ such that $\frac{a}{d}x+\frac{b}{d}y+\frac{c}{d}z=1$.
Let us assume that $ax+by+cz=e$ and $d\mid e$. Therefore, $ax+by+cz=dk$ for some $k\in\mathbb{Z}$, and $\frac{a}{d}x+\frac{b}{d}y+\frac{c}{d}z=k$. Since $k\in\mathbb{Z}$, we can instead solve the equation $\frac{a}{d}x+\frac{b}{d}y+\frac{c}{d}z=1$, and then multiply the resulting $x,y,z$ by $k$ to get the desired solution. Therefore, if $d\mid e$, then $ax+by+cz=e$ has a solution.
Let us now assume that $ax+by+cz=e$ and $d\nmid e$. Therefore, $\frac{e}{d}\notin\mathbb{Z}$. $\frac{a}{d}x+\frac{b}{d}y+\frac{c}{d}z=1$ has an integer solution for $x,y,z$, but this solution cannot be scaled up by $\frac{e}{d}$ since $\frac{e}{d}$ is not an integer. Therefore, if $d\nmid e$, $ax+by+cz=e$ does not have any integer solutions.

 A: The direction that you (correctly) had doubt about turns out to be the easy direction. 
We show that if the equation $ax+by+cz=e$ has a solution in integers, then the greatest common divisor $d$ of $a$, $b$, and $c$ must divide $e$. 
So suppose that $ax_0+by_0+cz_0=e$. Since $d$ divides each of $a$, $b$, and $c$, it follows that $d$ divides $ax_0+by_0+cz_0$, and therefore $d$ divides $e$.

The other direction is harder. Let $d$ be the greatest common divisor of $a$, $b$, and $c$. Suppose that $d$ divides $e$. Say $e=dk$. Then $a=da'$, $b=db'$, and $c=dc'$ for some integers $a',b',c'$ such that the greatest common divisor of $a'$, $b'$, and $c'$ is $1$.  You arrived at this stage, and quoted a theorem to the effect that there are integers $x$, $y$, and $z$ such that $a'x+b'y+c'z=1$.  Then 
$$e=(kd)(1)=(a'x+b'y+c'z=a(xk)+b(yk)+c(zk),$$
and the result follows.
This was, apart from minor differences of notation, your argument. It leaves a possible gap. It is a standard result that if the gcd of two integers $s$ and $t$ is equal to $1$, then there exist integers $x$ and $y$ such that $sx+ty=1$.
It is not quite so standard that if the gcd of three or more integers is equal to $1$, then some integer linear combination of the integers is equal to $1$. Let us prove the result, in our case of three integers $a,b,c$.
Let $m$ be the greatest common divisor of $a$ and $b$. Then $au+bv=m$ for some integers $u$ and $v$.  But the gcd of $a,b,c$ is $1$. It follows that $m$ and $c$ are relatively prime. So there exist  integers $w$ and $z$ such that $mw+cz=1$. 
Replacing $m$ by $au+bv$, we get $(au+bv)w+cz=1$, that is, $a(uw)+b(vw)+cz=1$. Thus $1$ is an integer linear combination of $a$, $b$, and $c$. 
A: I'll use some basic facts.


*

*The “greatest common divisor” operation on $\mathbb{N}$ is associative, that is
$$\gcd(a,\gcd(b,c)=\gcd(\gcd(a,b),c)$$
so we can denote this number by $\gcd(a,b,c)$.

*Every ideal in $\mathbb{Z}$ is of the form $a\mathbb{Z}$, for a unique $a\in\mathbb{N}$.

*If $I$ and $J$ are ideals in the ring $R$, then $I+J=\{x+y\mid x\in I, y\in J\}$ is an ideal of $R$. If $K$ is another ideal, then $I+(J+K)=(I+J)+K$.

*If $a,b\in\mathbb{N}$, then $a\mathbb{Z}+b\mathbb{Z}=c\mathbb{Z}$, where $c=\gcd(a,b)$.
Facts 1 and 3 are easily proved. Fact 2 is an application of the integer division with remainder and the fact that every non empty subset of $\mathbb{N}$ has minimum.
Let $I$ be an ideal of $\mathbb{Z}$. If $I=\{0\}$, then $I=0\mathbb{Z}$, so we can assume $I\ne\{0\}$. If $z\in I$, $z\ne0$, also $-z\in I$; thus the set of positive elements in $I$ is non empty. Let $a$ be its minimum ($a>0$ by hypothesis); in particular $a\mathbb{Z}\subseteq I$. Now, if $x\in I$, we can write $x=ay+r$, where $0\le r<a$. Since $x-ay\in I$, the case $r>0$ would contradict the minimality of $a$; therefore $r=0$ and so $x=ay\in a\mathbb{Z}$.
Let's tackle fact 4. We know from facts 2 and 3 that $a\mathbb{Z}+b\mathbb{Z}=c\mathbb{Z}$, for some $c\in\mathbb{N}$. Since $a\in c\mathbb{Z}$ because $a=a\cdot1+b\cdot0$, we have that $c\mid a$. Similarly $c\mid b$. If $d\mid a$ and $d\mid b$, then $a=dx$ and $b=dy$; but $c=ar+bs$ for some $r,s\in\mathbb{Z}$, so $c=dxr+dys=d(xr+ys)$ and $d\mid c$. Therefore $c=\gcd(a,b)$.
Now we can collect our facts: the set of integers that can be written as $ax+by+cz$ for $x,y,z\in\mathbb{Z}$ is just
$$a\mathbb{Z}+b\mathbb{Z}+c\mathbb{Z}=d\mathbb{Z},$$
where $d=\gcd(a,b,c)$. This shows that our set consists exactly of the multiples of $\gcd(a,b,c)$ and proves the claim.

Note. I'm not writing this to overwhelm with algebraic facts, but to show that going to abstract algebra can simplify the reasoning, by putting together some more general facts. For instance, the associativity of $\gcd$ can be seen as a consequence of the fact that $\mathbb{N}$ is a lattice under the “is a divisor of” order relation (but it's also provable directly).
One can compare a “pure” number theoretic proof with this one and find the common places.
