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edited Dec 4 2011 at 21:31

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.

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edited Dec 4 2011 at 21:10

Eric Naslund
3,870●14●34

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. This A proof of this is not difficult, at at the endof the question is a short proof.

My question is: What can be said Can we said about the number of zeros of $E_p(z)$ modulo $p$? What is improve the best bound for the number of zeros, and ? 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.

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asked Dec 4 2011 at 20:53

Eric Naslund
3,870●14●34

Truncated Exponential Series Modulo $p$: Deeper meaning for a Putnam Question.

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. This is not difficult, at at the end of the question is a short proof.

My question is: What can be said we said about the number of zeros of $E_p(z)$ modulo $p$? What is the best bound for the number of zeros, and is there a deeper connection 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.