Is $a$ or $b \equiv 0 \pmod{5}$ if $a^2-b^2 \equiv 0 \pmod{5}$?

Question

I have this question, knowing the last in title can i define that at least a or b is divisible by 5?

Answer 1

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.

Answer 2

Note that $0 = a^2-b^2 = (a+b)(a-b)$, so either $a-b \equiv 0 \pmod{5}$ or $a+b \equiv 0 \pmod{5}$

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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 $b$.

Answer 3

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.

Answer 4

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.