If $a^2 \equiv 4 \pmod{5}$, then $a \equiv 2 \pmod{5}$.

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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?

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2 Answers

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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 $a = 3\pmod 5$

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In general the implication is not true.

Note that $$5|(a^2-4) \implies 5|(a-2)(a+2) \implies 5|(a-2) \text { or } 5|(a+2)$$

Thus there are cases in which $5|(a^2-4)$ but it does not divide $(a-2)$.

For example $a=8$ is such a number $ a^2-4=60$ which is a multiple of $5$ but $a-2=6$ is not a multiple of $5.$

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