\section{On a Class of Elementary Symmetric Functions}%13

\markboth{Articles}{On a Class of Elementary Symmetric Functions}

\vspace{4.2cm}

\subsection{Introduction}

It is a well-known and classic result that if $p$ is a prime, then we have:

$$x^{p-1}-1\equiv (x-1)(x-2)\ldots (x-(p-1))\pmod p.$$

Hence the remainder of the elementary symmetric functions of the numbers

$1,2,\ldots $, $p-1\pmod p$

are zero, except for

$\sigma _{p-1}$.

We can obtain that their product is $\equiv -1\pmod p$ which is Wilson's theorem.

Moreover, Wolstenholme proved that the numerator of the fraction

$$\ds\f{1}{1}+\ds\f{1}{2}+\ds\f{1}{3}+\ldots +\ds\f{1}{p-1}$$

when written in lowest terms is a multiple of $p^2$ for every prime number $p\ge 5$.

Despite Wolstenholme's result, this type of problems keeps appearing even

in olympiads.

For example in the XX Iberoamerican Olympiad held in Cartagena,

Colombia, one of the problems was to prove that the numerator of the

fraction

$$\ds\f{1}{1^p}+\ds\f{1}{2^p}+\ds\f{1}{3^p}+\ldots +\ds\f{1}{(p-1)^p},$$

when written in lowest terms, was divisible by $p^3$ for every $p\ge 5$.

The purpose of this article is not only to prove these results, but also to

provide a solution for a more general case, showing some ideas that might be

useful for this kind of problems.

Of course all the discussed problems have

elementary solutions; however, the techniques used for them do not work in a

more general case, and we are forced to use stronger artillery.

It is time to do some mathematics before the reader gets bored of all this

talking.

\subsection{The Result}

Before continuing, let's first introduce the theorem we are going to prove.

{\bf Theorem 1.}

{\it Let $p$ be a prime number greater than 3 and $k$ a positive integer.

Let

\begin{align*}

P(x)

& =(x-1^{p^k})(x-2^{p^k})(x-3^{p^k})\ldots (x-(p-1)^{p^k})\\

& =x^{p-1}-x^{p-2}A_{(1,k)}+x^{p-3}A_{(2,k)}-\ldots -xA_{(p-2,k)}+A_{(p-1,k)}.

\end{align*}

Then the following statements are true:

1. $A_{(i,k)}$ is divisible by $p^{k+1}$ for $1\le i\le p-3$.

2. $A_{(p-2,k)}$ is divisible by $p^{k+2}$.

3. $A_{(p-1,k)}+1$ is divisible by $p^{k+1}$.

}

Before proving this result we must prove the following lemma.

{\bf Lemma 1.}

(Lifting the exponent).

{\it Let $p$ be an odd prime.

If $a\equiv b\pmod {p^t}$ then $a^p\equiv b^p\pmod {p^{t+1}}$.

}

{\bf Proof.}

Let $a=b+mp^t$, therefore

$$a^p=b^p+\sum_{k=1}^p \ds\binom{p}{k}b^{p-k}m^k p^{tk}.$$

Looking at this relation $\pmod {p^{t+1}}$ and since

$p\mid \ds\binom{p}{k}$ for $k=1,2,\ldots ,p-1$,

it follows that in fact $a^p\equiv b^p\pmod {p^{t+1}}$, and this completes the proof of the

lemma.

Now we can continue, but for those who thought we were only going to use

classical number theory arguments, you were wrong!

Instead, we try a more

algebraic approach by considering the polynomial

\begin{align*}

Q(x,y)

& =(x-y)(x-y^2)\ldots (x-y^{p-1})\\

& =x^{p-}+x^{p-2}f_1(y)+x^{p-3}f_2(y)+\ldots +xf_{p-2}(y)+f_{p-1}(y).

\end{align*}

We will have to work a little with this polynomial, and the reason we picked it

will be clear later.

The first thing we are going to prove is that $\phi _{p-1}(x)$ divides

$f_i(x)$ for $i=1,2,\ldots ,p-2$ in $\mathbb{Z}[x]$, where $\phi _{p-1}(x)$

is the $(p-1)th$

cyclotomic polynomial.

This follows easily by plugging $y=\zeta $ where $\zeta $ is a $(p-1)$th primitive root of

unity.

For this value we get

$Q(x,\zeta )=x^{p-1}-1$

so

$f_i(\zeta )=0$ for $i=1,2,\ldots ,p-2$

because the latter is a polynomial identity in $x$.

Since the roots of $\phi _{p-1}(x)$

are precisely the $(p-1)$th primitive roots of unity, and

$\phi _{p-1}(x)\in \mathbb{Z}[x]$,

we get the desired divisibility.

Now it is known that a primitive root mod $p^2$ is also a primitive root mod

$p^t$ for every positive integer $t$.

Let $g$ be a primitive root mod $p^t$ for all $t$ (just pick one mod $p^2$ and it works),

then

$$p^{k+1}\mid \phi _{p-1}(g^{p^k}).$$

This is not hard at all, since

$p^{k+1}\mid (g^{p^k})^{p-1}-1$,

$p^{k+1}$ must divide some cyclotomic polynomials that are divisors of $x^{p-1}-1$

evaluated at $g^{p^k}$.

Suppose

$p\mid \phi _d(g^{p^k})$ then

$p\mid g^{p^k d}-1$

and since the order of $g$ mod $p$ is $(p-1)$ and $d\mid p-1$,

we must have $d=p-1$,

proving that all factors of $p$ must divide $\phi _{p-1}(g^{p^k})$,

and hence we get the desired divisibility.

Now, using the last result it is easy to finish the proof of the first part of the theorem.

The numbers $g^{1p^k},g^{2p^k},\ldots ,g^{(p-1)p^k}$

are in some order the remainders

$1^{p^k},2^{p^k},3^{p^k},\ldots ,(p-1)^{p^k}$ mod $p^{k+1}$

because of the lifting lemma.

Now consider

\begin{align*}

Q(x,g^{p^k})

& =(x-g^{p^k})(x-g^{2p^k}\ldots (x-g^{(p-1)p^k})\\

& =x^{p-1}+x^{p-2}f_1(g^{p^k})+\ldots +xf_{p-2}(g^{p^k})+f_{p-1}(g^{p^k}).

\end{align*}

By the previous remarks we now have that

$f_i(g^{p^k})\equiv 0\pmod {p^{k+1}}$, for $i=1,2,\ldots ,p-2$, but

$f_i(g^{p^k})\equiv A_{(i,k)} \pmod {p^{k+1}}$,

also from the previous remarks, thus the first part of the theorem follows.

In order to prove the second part we must deal with a complicated divisibility

relation first; we prove it here as a lemma.

{\bf Lemma 2.}

{\it Let $p$ be an odd prime and $k$ a positive integer, then for every

integer $a$ not divisible by $p$ we have

$$p^{2k+3}\mid p^{2k+2}-p^{k+1}(a^{p^k}+(p-a)^{p^k}).$$

}

{\bf Proof.}

The divisibility is equivalent to

$p^{k+2}\mid p^{k+1}-(a^{p^k}+(p-a)^{p^k})$

and we will prove it by induction on $k$ using the lifting lemma.

The cases $k=0$, $k=1$

are directly verified and are easily reduced to Fermat's little theorem.

The inductive step is not easy at all:

$$p^{k+1}=a^{p^k}+(p-a)^{p^k}\pmod {p^{k+2}}$$

$$p^{k+1}-a^{p^k}=(p-a)^{p^k}\pmod {p^{k+2}}$$

$$(p^{k+1}-a^{p^k})=(p-a)^{p^{k+1}}\pmod {p^{k+3}}$$

$$p^{k+2}a^{(p-1)p^k}-a^{p^{k+1}}=(p-a)^{p^{k+1}}\pmod {p^{k+3}}$$

$$p^{k+2}=a^{p^{k+1}}+(p-a)^{p^{k+1}}\pmod {p^{k+3}}.$$

The third step is due to the binomial formula and the second one is by the

lifting lemma.

And that finishes the proof of the lemma.

To prove the second part of the theorem we will prove that the difference

$P(p^{k+1})-A_{(p-1,k)}$

is divisible by $p^{2k+3}$.

Multiplying and using Lemma 2,

$$(p^{k+1}-a^{p^k})(p^{k+1}-(p-a)^{p^k})\equiv a^{p^k}(p-a)^{p^k}\pmod {p^{2k+3}},$$

and multiplying this congruence for

$a=1,2,\ldots ,(p-1)/2$

we obtain the stated divisibility,

since

$A_{(p-1,k)}=\ds\prod_{i=1}^{p-1}i^{p^k}$.

We can also expand the difference as

$$P(p^{k+1})-A_{(p-1,k)}=p^{(k+1)(p-1)}-p^{(k+1)(p-2)}A_{(1,k)}$$

$$+p^{(k+1)(p-3)}A_{(2,k)}\ldots -p^{k+1}A_{(p-2,k)},$$

and since $p^{2k+3}$ divides all terms but the last one in that sum (by the first

part of the theorem and a simple $p$-adic valuation), we get that $p^{2k+3}$ must

divide the last term.

But this in turn implies that $p^{k+2}\mid A_{(p-2,k)}$ as wished.

Note that Wolstenholme's theorem and the Iberoamerican Olympiad problem

follow from this part of the theorem with $k = 0$, $k = 1$ respectively.

To finish with the third part of the theorem we only have to take Wilson's

congruence and apply the lifting $k$ times.

Doing this we get

$1^{p^k}2^{p^k}\ldots (p-1)^{p^k}\equiv (-1)^{p^k}\pmod {p^{k+1}}$,

which is exactly the third part of the theorem,

and this concludes our proof.

\bigskip

\hfill

{\Large Pascual R. Mesa, Colombia}