If c is an odd integer, then:
$$c = 2k + 1 $$
where \(k\) is an integer. This means:
$$c^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1$$
Rearranging:
$$c^2 -1 = 4(k^2 + k) $$
Since \(k^2+k\) is an integer, then \(c^2 ≡ 1 \bmod 4\). Also, \(k^2+k\) can be written as \((k)(k+1)\). Since either \(k\) or \(k+1\) is even, then \((k) (k + 1) = 2q\). In other words:
$$\begin{align} c^2 -1 &= 4(k^2 + k) \\ &= 8q \end{align} $$
Therefore \(c^2 ≡ 1 \bmod 8\).