<s>[INST] show that if r and s are odd, then ${(rs)^2 -1\over 8} \equiv {r^2 -1\over 8} + {s^2 -1\over 8}$ mod 8. show that if r and s are odd, then ${(rs)^2 -1\over 8} \equiv {r^2 -1\over 8} + {s^2 -1\over 8}$ mod 8.
This was part C of the problem. I was able to solve A and B, so I can use the following facts:
A. ${rs -1\over 2} \equiv {r -1\over 2} + {s -1\over 2}$ mod 2
B. $r^2 \equiv s^2 \equiv 1$ mod 8
 [/INST] Showing that
$$\frac{(rs)^2-1}{8}\equiv \frac{r^2-1}{8} + \frac{s^2-1}{8} \mod 8$$
is equivalent to showing that
$$\frac{(rs)^2-1}{8} - \frac{r^2-1}{8} - \frac{s^2-1}{8} \equiv 0 \mod 8$$
which is equivalent to
$$\frac{r^2s^2-r^2-s^2+1}{8} \equiv 0 \mod 8$$
or
$$\frac{(r^2-1)(s^2-1)}{8} \equiv 0 \mod 8$$
which is true since both terms in the product in the numerator are divisible by $8$.
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