Show that $a^2 \not\equiv b^2 \pmod p$ Given that: $p$ is an odd prime, $a\geq 1,b \leq \frac{p-1}{2}$, and that $a \not\equiv b \pmod p$, show that $a^2 \not\equiv b^2 \pmod p$
Based on the second condition, I know that $a + b < p$ is true, which I think should be helpful, but I'm not sure how to use this.
I also tried to use the fact that $p = 2k + 1$, but again, I didn't find any luck with this.
I'm very new with congruences, so I really appreciate any help!
 A: If $p\mid a^2-b^2=(a-b)(a+b)$, then $p\mid a+b$ or $p\mid a-b$ because it is prime. Now as you correctly noted $a+b\lt p$, so $p$ cannot divide $a+b$, therefore $p\mid a-b$; in other words $a\equiv b\pmod{p}$.
A: If 
$a^2 \equiv b^2 \mod p \tag 1$
then
$a^2 - b^2 \equiv 0 \mod p, \tag 2$
or
$(a + b)(a - b) \equiv a^2 - b^2 \equiv 0 \mod p; \tag 3$
since
$1 \le a, b \le \dfrac{p - 1}{2}, \tag 4$
we have
$2 \le a + b \le 2\dfrac{p - 1}{2} = p - 1; \tag 5$
it follows that 
$a + b \not \equiv 0 \mod p, \tag 6$
and hence that $a + b$ is invertible modulo $p$, since
$\Bbb Z_p = \Bbb Z / p\Bbb Z \tag 7$
forms a field, and $p$ is odd, so $2 \not \equiv 0 \mod p$; from this we conclude that
$a - b \equiv 0 \mod p, \tag 8$
or
$a \equiv b \mod p,  \tag 9$
contrary to the hypothesis
$a \not \equiv b \mod p; \tag{10}$
therefore,
$a^2 \not \equiv b^2 \mod p, \tag{11}$
$OE\Delta$.
A: The idea is that $a^2-b^2=(a-b)(a+b)$ and saying $a^2 \not\equiv b^2 (mod p)$ is the same as saying, $(a-b)(a+b) \not\equiv 0 (mod p)$
You do have from your conditions that $a-b \not\equiv 0 (mod p)$ and since $a+b<p$ then ofcourse $a+b \not\equiv 0 (mod p)$ and $p$ is a prime so now you have that, $a^2 \not\equiv b^2 (mod p)$
