elementary number theory - If $a^2 equiv 4 pmod{5}$, then $a equiv 2 pmod{5}$.

I have the following proof to complete:

If $a^2 \equiv 4 \pmod{5}$, then $a \equiv 2 \pmod{5}$ for all $a \in \mathbb{Z}$.

Here is what I have so far:

Because $a^2 \equiv 4 \pmod{5}$, we have $5 \mid (4-a^2)$, and thus there is $c \in \mathbb{Z}$ s.t. $(4-a^2)=5c$. We can manipulate this equation in order to get $a^2=4-5c$.

I see how taking the square root of the $4$ will give me my answer, but I'm not sure what to do with the $5c$. I realize that this is equivalent to $0 \ ( \text{mod } 5)$, so can I just say
$4-5c \equiv 4-0 \ (\text{mod } 5)$, so $a \equiv 2 \pmod{5}$?

It feels like that just brings me back to my starting point, and I don't think that radicals are allowed under modulo. So what do I do?

Answer

We have $$ a=0,1,2,3,4 \pmod 5$$
so $$a^2 = 0,1,4,4,1\pmod 5$$ so it is not neccesarly to have $a=2 \pmod 5$

Example, if $a=3$ then $a^2 = 4\pmod 5$ but $a = 3\pmod 5$