What constitutes a proof of congruence in modular arithmetic? 
In this problem, don’t use a calculator. The answers can be derived without doing much computation, try to find these simple solutions.
(a) $4 + 5 + 6 ≡ 0 \pmod {5}$

My professor gave me some congruences to prove in my latest discrete HW. They seem pretty easy, but I'm not sure what constitutes a proof. I know that a congruence of numbers $a, b$ and $n$ would have the following characteristics:
1) $a-b$ would be a multiple of $n$
2) $a$ is $b$ more than a multiple of $n$
3) $a / n$ has remainder $b$
...would showing any of these three things to be true (or all three to be true) constitute a proof? 
 A: Better: $5\equiv 0\pmod 5$, $6\equiv 1\pmod 5$, and $4\equiv -1\pmod 5$, so $4+5+6\equiv -1+0+1 \equiv 0\pmod 5$.
A: We have that $4 + 5 + 6 = 15$. Recall the definition of the congruence relation: $a \equiv b \pmod{n}$ if and only if $n \big | (a - b)$. So $5 \big | (15 - 0)$, which implies that $4 + 5 + 6 \equiv 0 \pmod{5}$. 
A: All three of those imply $a\equiv b(mod~n)$.
1) $n|(a-b)$ 
By the definition of modular equivalence, $a\equiv b(mod~n)$
2) $a=kn+b$ for some $k\in \mathbb{Z}$
Moving the b to the LHS reveals $a-b=kn\Rightarrow n|(a-b)\Rightarrow a\equiv b(mod~n)$
3) "$a/n$ leaves remainder $b$" is the same as "$a=kn+b$ with $0\leq b<n$". This, as in 2), implies $a\equiv b(mod~n)$.
It should be noted that for 3), the converse isn't true, i.e. $a\equiv b(mod~n)$ does NOT imply "$a/n$ leaves remainder $b$". As a counterexample, $5\equiv 3(mod~2)$ but 5 leaves remainder 1, not 3, after division by 2.
A: I would add that you can certainly use, as a variation on 1) in your question, 
$$a+kn \equiv a \bmod n$$
... effectively, "casting out $n$s", which can be used to good effect in your example question:
\begin{align}
4+5+6 &\equiv 4+6 \bmod 5\\
&\equiv (5-1)+(5+1) \bmod 5\\
&\equiv -1 + 1 \bmod 5  \\
&\equiv 0 \bmod 5\\
\end{align}
or, if using negative numbers in modulus calculations is not yet "approved",
\begin{align}
4+5+6 &\equiv 4+6 \bmod 5\\
&\equiv 4+(5+1) \bmod 5\\
&\equiv 4 + 1 \bmod 5  \\
&\equiv 5 \bmod 5\\
&\equiv 0 \bmod 5\\
\end{align}
