1. Let $p$ be a positive odd number. Then the remainder of $p^{2}$ divided by 8 is $\qquad$ .
Because $p$ is an odd positive integer, let $p=2k-1\left(k \in \mathbf{N}_{+}\right)$, so
$$
p^{2}=(2k-1)^{2}=4k^{2}-4k+1=4(k-1)k+1 \text{. }
$$

Since $(k-1)k$ is even, therefore, $p^{2}$ leaves a remainder of 1 when divided by 8.