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Extension of Daniel Loughan's example (which is also known, but not as well-known as it should(?) be): if a prime $p$ is of the form $dn+1$ then the Fermat hypersurface $\sum_{i=1}^d x_i^d = 0$ in ${\bf P}^{d-1}({\bf F}_q)$ is smooth and its number of rational points is not congruent to $1 \bmod p$. Indeed the usual argument for Chevalley(-Warning) shows that the number of rational points is congruent mod $p$ to $1 \pm t$ where $t$ is the $(x_1 x_2 \cdots x_d)^{p-1}$ coefficient of $\left(\sum_{i=1}^d x_i^d\right)^{p-1}$, and when $p = dn+1$ this coefficient is $(p-1)!/n!^d$ which is clearly not $0 \bmod p$.