Is $a$ or $b \equiv 0 \pmod{5}$ if $a^2-b^2 \equiv 0 \pmod{5}$? I have this question, knowing the last in title can i define that at least a or b is divisible by 5?
 A: Note that $0 = a^2-b^2 = (a+b)(a-b)$, so either $a-b \equiv 0 \pmod{5}$ or $a+b \equiv 0 \pmod{5}$

UPDATE
So you are saying $a^2 \pmod{5} = b^2 \pmod{5} = 4$. If $a$ is divisible by $5$, so is $a^2$, so since $a^2 \pmod{5} = 4 \ne 0$, you see that $a$ is not divisible by $5$. Similarly, neither is $b$.
A: Noting your comment to gt6989b's answer, use that
$$a^2-4\equiv 0 \pmod 5 \implies 5|(a-2) \text{ or }5|(a+2)$$
Because $a^2-4=(a-2)(a+2)$.
For both of these, $5|a$ is impossible.
A: Well, ... no because $a^2 - a^2 \equiv 0$ but neither $a$ need not be equivalent to $0 \mod 5$.
Or for that matter $a^2 - (-a)^2 \equiv 0$ but if $a \ne \equiv 0 \pmod 5$ then $-a\equiv 5-a\not \equiv 0$ either.
You can exhaust all cases.  $0^2 - 0^2, 1^2 - 1^2, 1^2 - 4^2, 2^2 - 2^2, 2^2 - 3^2, 3^2 - 3^2, 3^2 - 2^2, 4^2 - 4^2, 4^2 - 1^2$ are all equiv $0$.
In fact we can see that if $a^2 - b^2 \equiv 0$ then either both are equivalent to zero or neither are.
Which makes perfect sense becuase if one of them, say $b$, was equivalent to $0$ then $a^2 - b^2 \equiv a^2 - 0^2 \equiv a^2\pmod 5$ (Or if $a\equiv 0$ then $a^2 - b^2 \equiv 0^2 - b^2 \equiv -b^2 \equiv 0\pmod 5$) and there's no reason to assume the other squared (or negative the other squared) would be $0$.
If we think about it is seems really weird that one would think it would imply one is equiv to $0$.
Why DID you think that?
...
Note also $a^2 - b^2 = (a+b)(a-b)\equiv 0 \pmod 5$.  Because $5$ is prime that means either $5|a+b$ and $a+b \equiv 0 \pmod 5$ or $5|a-b$ and $a-b \equiv 0 \pmod 5$.  Those would mean that if $a^2 - b^2 \equiv 0\pmod 5$ we MUST have either $a \equiv b\pmod 5$ or $a \equiv -b \pmod 5$.  
That is almost the exact opposite of your assumption.
However it's worth noting that $a^2 -b^2 \equiv 0 \pmod n$ wont mean this if $n$ is not prime.
A: Let's supposed that $ a \equiv 0 \pmod{5} $, so $a^{2} \equiv 0 \pmod{5}$, as consequence $b \equiv 0 \pmod{5}$, would also be true, if $a^{2} - b^{2} \equiv 0 \pmod{5}$. 
So, if one of the two is multiple of 5, then the other also has to be and if one is not multiple of five, then the other also cannot be.
