How to prove or disprove that if $x \equiv c \mod n$ then $x \equiv -c \mod n$

I am studying modular arithmetic and I don't know how to prove or disprove the following :

If $x \equiv c \mod n$ then $x \equiv -c \mod n$

By trying different numbers it seems as this is true ,some attempt I've made so far is :

$x \equiv c \mod n \implies x \equiv -(-c) \mod n$

Since $-x \equiv -c \mod n$ I have that $$x \equiv -(-x) \mod n \implies 0\equiv 0 \mod n$$

What does the last statement now mean ?

• Did you try some specific numbers? It should be fairly clear, either from the definition of $\equiv$ or the final step you have that this is false. – user296602 Jan 23 '16 at 17:48
• I've picked up the wrong numbers then ... – Nameless Jan 23 '16 at 17:49
• You must have only picked $n=2$ or $x=nm$ for it to be true. – user208649 Jan 23 '16 at 17:50
• Try doing this with $n=3$; what does that give you? – Arnaud D. Jan 23 '16 at 17:50
• Before the edit, you had come to the conclusion that $2x \equiv 0 \pmod n$. Write down any number for which this is false. – user296602 Jan 23 '16 at 17:50

Mod 2, $1\equiv-1$ and the $\iff$ is true. When the mod $n$ is $>2$, $1\not\equiv-1$ and the $\iff$ is false.