TITLE: A question on maps from $\mathbb{Z}/p\mathbb{Z}$ to itself
QUESTION [37 upvotes]: Let $p\geq 3$ be a prime number, and let $u:\mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}$ be a map such that, for all $l\in \mathbb{Z}/p\mathbb{Z}$,$l\neq 0$, the map $k\mapsto u(k+l)-u(k)$ is a permutation. Is $u$ a polynomial of degree $2$? 
Note that the property clearly holds when $u$ is a polynomial of degree $2$. Explicit computations seem to show that the converse holds -- that is, the answer is positive -- for $p$ at most $13$.
This is (in a non-obvious way) a special case of this other question, but presumably the statement here is much easier. The question came up quite naturally when thinking about some aspects of complex Hadamard matrices.

REPLY [24 votes]: The question was an open problem on planar functions for many years, which was settled independently in three papers around 1990. See the papers by Gluck, Ronyai and Szonyi, and Hiramine. Elkies' answer is similar to Gluck's proof. The proof by Hiramine avoids Segre's theorem, it is based on quite complicated computations instead.