Modulo multiplicative inverse

In ax $$\equiv$$ 1 (mod m) , when gcd(a, m) = 1, there is exactly one solution, i.e., when it exists, a modular multiplicative inverse is unique.

This is written in wikipedia. I am confused because i keep on thinking if a = 2 and m = 3. Then x can be 2,5,8 etc. I know i am missing something but i am unable to get it. I am on my own and thus would be happy if someone can clear my doubt.

• That solutions of a congruence equation are unique means any two solutions are congruent (vs. equal). – Bill Dubuque May 15 at 15:56
• Thanks, I understand totally. – Prasoon Mishra May 16 at 3:37

The important idea here is that we are working in the ring $$\mathbb{Z}/m$$, and questions about uniqueness of inverses should be asked in this context. Taking the $$m$$ elements of $$\mathbb{Z}/m$$, there is only one of them that's an inverse to any given $$x$$.
In your example, all of $$2$$, $$5$$, $$8$$, $$11$$, $$14$$ and so on are inverses to $$2 (mod 3)$$. But they're all equal to $$2 (mod 3)$$, so as far as $$\mathbb{Z}/3$$ is concerned that's the only inverse.
If we have $$a=2$$ and $$m=3$$ then we have the equation $$2x=1\mod3$$, there are multiple solutions to this notably $$x=3n+2$$ for $$n \in \mathbb{Z}$$.
Just an edit: Even though there are lots of solutions, there will only be one solution for $$\mod3$$ and in $$\mod3$$ there are the numbers $$\{0,1,2\}$$. Where the only solution to this is $$2$$.