Modular Arithmetic Simple question Hi I am studying modular arithmetic just for myself, but to be honest I find it very difficult. I am not sure which answer is right. Is it right to say 
If $4x \equiv 8\pmod{15}$ then $x \equiv 2 \pmod{15}$ or just $x=2$ ?
Thank you
 A: The question is not well-formed: in modular arithmetic,
$$
a \equiv b \mod(n) \iff n \text{ divides } a - b
$$
In your case, you are not using any number/entity as "$b$" in the first equation, so the equations you wrote are incomplete, so the question does not have sense. 
A: Hint: See what happens when $x = 17$, for example.
The reason you can deduce $x\equiv 2\pmod{15}$ from $4x \equiv 8\pmod{15}$ is that $4\times 4 \equiv 1 \pmod{15}$. If $4x\equiv 8 \pmod{15}$ then also $16x\equiv 32\pmod{15}$ (multiplying by $4$: this is legal since if $15|4x-8$ then also $15|4(4x-8)$). However, $16\equiv 1\pmod{15}$ and $32\equiv 2\pmod{15}$, this implies $x\equiv 2\pmod{15}$ (this is a bit harder to justify, but hopefully you saw this in class).
We can say a little more: if $x \equiv 2\pmod{15}$ then multiplying by $4$, we get $4x\equiv 8\pmod{15}$. So both statements are equivalent. This means that if you take any $x$ satisfying $x\equiv 2\pmod{15}$ then you will have $4x \equiv 8\pmod{15}$.
A: $$4x\equiv 8\pmod{15}\iff x\equiv 2\pmod{15}.$$
This is true because $4$ (by which you divide the both sides) and $15$ are coprime. (otherwise you cannot divide the both sides.) But note that $x=2$ is not the only solution. The answer is $x=15k+2$ where $k\in\mathbb Z$. For example, $x=17,32,47,\cdots$ satisfy $x\equiv 2\pmod{15}$.
A: Hopefully it helps to make a table
n | 4n mod 15
0 | 0
1 | 4
2 | 8
3 | 12
4 | 1
5 | 5
6 | 9
7 | 13
8 | 2
9 | 6
10| 10
11| 14
12| 3
13| 7
14| 11
So there is a one to one correspondence between n in {0..14} and 4*n mod 15.
x is congruent to 2 mod 15 is different from x = 15 because the first one includes 17.
Regards,
Matt
