Proof of divisibility: if $a|b$ and $a|(b+c)$ then $a|c$ This one is killing me, any help is greatly appreciated!
 A: Go back to the definition of divisibility: $a\mid b$ means that there is an integer $m$ such that $b=ma$, and $a\mid b+c$ means that there is an integer $n$ such that $b+c=na$. You’re interested in $c$, so isolate it:
$$c=na-b=na-ma=(n-m)a\;.$$
Is $n-m$ an integer? Does this show that $a\mid c$?
A: Hint $\rm\ \ \dfrac{b+c}a,\,\dfrac{b}a\in\Bbb Z\ \Rightarrow\ \dfrac{c}a = \dfrac{b+c}a-\dfrac{b}a\in\Bbb Z\ $ since $\,\Bbb Z\,$ is closed under subtraction.
Remark $ $ More generally, sets of common multiples are closed under subtraction, so closed under mod (= repeated subtraction), so closed under gcd (= repeated mod), which is at the heart of the Fundamental Theorem of Arithmetic (more general: Euclidean $\Rightarrow$ PID $\Rightarrow$ UFD for domains).
A: If $a\mid b$ and if  $a\mid (b+c)$, there exist  some integers $k,l$
such that $b=k\cdot a,b+c=l\cdot a $
So,$c=b+c-b=la-ka=a(l-k)\implies \frac c a=k-l$ some integer
A: If $a|b$ then $b=a\cdot n$.
If $a|(b+c)$ then $b+c=a\cdot m$.
Hence $c=b+c-b=a\cdot m-a\cdot n=a\cdot(m-n)$, so $a|c$
A: In general, if $a\mid m$ and $a\mid n$ then $a$ divides any linear combination of $n$ and $m$. That is, for all $x,\ y\in\mathbb{Z}$, we have $a\mid mx + ny$. 
Given these facts, can you now find a linear combination of $b$ and $b+c$ which gives $c$?
A: $a|b$ mean that exists whole number $k$ such that  $$b=ka \dots(1)$$
and $a|(b+c)$ mean that exists whole number $l$ such that $$(b+c)=la\dots(2)$$
replacing (1) in (2) we get
$$ka+c=la$$
$$c=la-ka$$
$$c=a(l-k)$$
because $l-k=r$ is whole number that mean$$c|a$$
A: Note that, by definition of divisibility, $a|b$ implies
$$b = ak$$
for some integer $k$. Also, we have that $a|(b + c)$ implies
$$b + c = al$$
for some integer $l$. So, substitute the first into the second and solve for c. You should get $c = a(l - k)$, which implies $a|c$.
A: (In this answer all variables are integers, i.e., elements of $\mathbb{Z}$.)
By the definition of divisibility we are given that $\;n * a = b\;$ and $\;m * a = b+c\;$ for some $\;n\;$ and $\;m\;$.  Now we are asked to find a $\;k\;$ which makes $\;k*a = c\;$:
\begin{align}
& k*a = c \\
\equiv & \;\;\;\;\;\text{"use the only fact we know about $\;c\;$"} \\
& k*a = m*a - b \\
\equiv & \;\;\;\;\;\text{"use the other fact"} \\
& k*a = m*a - n*a \\
\equiv & \;\;\;\;\;\text{"factor out $\;a\;$ -- to make both sides more alike"} \\
& k*a = (m-n)*a \\
\Leftarrow & \;\;\;\;\;\text{"weaken using Leibniz' rule -- to achieve our goal"} \\
& k = m-n \\
\end{align}
Therefore we have found such a $\;k\;$, and hence proved $\;a|c\;$.
(Yes, this answer looks a lot like https://math.stackexchange.com/a/452159/11994 since both questions are a lot alike.  Apologies if the duplication is against site policy.)
