Let $X$ be a complete intersection in $\mathbb{P}^n$ of multidegree $(d\_1,\ldots,d\_r)$. If we're working over a finite field $\mathbb{F}\_q$, the Ax-Chevalley-Warning theorem says that if $X$ is in the Fano range, i.e., $$ \sum d\_i \leq n,$$ then $|X(\mathbb{F}\_q)| \equiv 1 \pmod q.$


In the case that $X$ is not in the Fano range, one can cook up examples of such $X$ with no $\mathbb{F}\_q$-points at all using the norm form, but these are not smooth.


In SGA 7 II Exposè XXI, Katz shows that for a general complete intersection in the Fano range, this congruence is not satisfied, possibly after extending the ground field.


Given a multidegree outside the Fano range, are there *explicit* examples of *smooth* complete intersections which don't satisfy this congruence?


Edit: Sorry, I wasn't sufficiently explicit about what I was asking. Restricting to hypersurfaces, what I would like is for each $d$, $p$, and $n$ with $d \geq n+1$, an explicit example of a smooth hypersurface of degree $d$ in $\mathbb{P}^n$ which doesn't satisfy the Chevalley-Warning congruence.


Thanks for the examples, though; I knew some explicit examples before but no infinite families.



Smooth complete intersections and sharpness of the Chevalley-Warning theorem
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 $0 \bmod p$. 


