Apparently B6 of the Putnam this year asked:
Suppose $p$ is an odd prime. Prove that for $n\in \{0,1,2…p-1\}$, at least $\frac{p+1}{2}$ of the numbers $\sum^{p-1}_{k=0} k! n^{k}$ are not divisble by $p$.
With some rearrangements, this is equivalent to showing that $$E_p(z):=\sum_{k=0}^{p-1} \frac{z^k}{k!}$$ has at most $\frac{p-1}{2}$ zeros. A proof of this is at the end.
My question is: Can we improve the bound for the number of zeros? Also is there a deeper connection here with other parts of mathematics motivating this problem?
Proof of problem: Consider $$Q(z)=z^{p}-z+\sum_{l=0}^{p-1}\frac{z^{l}}{l!}.$$ Then for each integer $Q(n)=E(n).$ However, $$Q^{'}(z)\equiv E^{'}(z)-1=E(z)-\frac{z^{p-1}}{(p-1)!}-1\equiv E(z)+z^{p-1}-1.$$ Then, if $Q(n)=0$ for $n\neq0$ , we must also have $Q^{'}(n)=0$ so that $n$ is a double root of $Q(n).$ Since $\deg Q(n)=p$, we see that at most half of the integers $n\in\{ 1,2,\dots,p-1\}$ satisfy $E(n)=0.$ Since $E(0)=1$, we conclude the desired result.
Remark: This was asked in a slightly different form on math stack exchange. I felt the answer I posted there was inadequate there, and I personally became more curious while attempting to answer the question.
Best Answer
I've seen this trick in a paper of Mit'kin, Math Zametki 1992. There he improves the bound. This is related to Stepanov's method to bound the number of solutions of equations over finite fields.