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\centerline{\bf On a question of Buium }
\medskip
\centerline{\bf Jos\'e Felipe Voloch}
\bs

For an integer $n$ and a prime $p$, the quantity $\d_p(n)=(n^p-n)/p (\mod p)$,
has been considered classically. In fact $\d_p(n)/n$ is known as the Fermat
quotient. Recently this quantity has been reconsidered as part of the quest
for finding a substitute, in the number field field case, for the derivations
in the function field case (see [B1,2],[I],[Sm]), since it satisfies the
Leibniz rule, that is $\d_p(mn)= m\d_p(n)+n\d_p(m)$.

Let $R$ be the ring $\prod_p \F_p$, where the product is taken over all
primes, then $R$ is a ring of characteristic zero
(not a domain) and the integers $\Z$ sit in $R$. Also, given an integer $n$,
$\d(n)=(\d_p(n))_p$ is an element of $R$. Buium asked the following question:
decide if $\d(n)$ is in $\Z$ for all $n$. Clearly $\d(n)=0, n=0,\pm 1$.
If there are infinitely many Mersenne primes we will show that $1,\d(2)$ are
linearly independent over $\Z$. 
Assuming a generalization of this conjecture, we will prove more.
Namely, if $n_1,...,n_r$ are multiplicatively independent integers then
$1,\d(n_1),\ldots,\d(n_r)$ are linearly independent over $\Z$.

Consider the following statements:

(A) If $m>n\ge 1$ are coprime integers, such that $m/n$ is not a perfect power,
then there are infinitely many primes of the form $(m^l-n^l)/(m-n)$.

(B) If $m,n\ne 0$ are integers, $m/n \ne \pm 1$, then $1,m\d(n)-n\d(m)$ are 
linearly independent over $\Z$.

(C) If $n_1,...,n_r$ are multiplicatively independent non-zero integers then
$1,\d(n_1),\ldots,\d(n_r)$ are linearly independent over $\Z$.

The statement (A), at least when $n=1$, is a well-known open problem 
in elementary number theory and it
is widely believed to be true, although no cases of it has been proved.
The special case $m=2,n=1$ corresponds to Mersenne primes and there there is
ample numerical evidence. The case $m=10,n=1$ corresponds to the so-called
repunits and there there is also some numerical evidence. The 
statement (B), with $n=1$, is an answer to Buium's question, 
while (C) generalizes (B). We prove:

\proclaim Theorem. (A) implies (B) and (B) implies (C).

{Proof:} Assume $m,n$ are integers as in (A) and assume (A) holds.
Let $p=(m^l-n^l)/(m-n)$ be prime. Then $m^l = n^l + p(m-n)$.
If $l$ does not divide $p-1$, then $x \mapsto x^l$ is a bijection
in $\Z/p$ and from $m^l \con n^l (\mod p)$, we conclude that $p|(m-n)$
which will be false for $p$ large.
Assume that is not the case, so that $l|(p-1)$. Then
$$m^{p-1} = (n^l + p(m-n))^{(p-1)/l} \con n^{p-1} -{(m-n)p\over ln^l}
(\mod p^2).$$
Thus,
$$n\d_p(m)-m\d_p(n) \con -{(m-n)nm\over ln^l} (\mod p).$$
 
If (B) is false, there exists $a,b$ integers not both zero with
$a(n\d(m)-m\d(n))+b=0$ so, for $p$ as above we get
$aln^l-bnm(m-n) \con 0 (\mod p)$. For $p$ going to infinity
of the form $(m^l-n^l)/(m-n)$ we have $ln^l = o(p)$, since $m > n$.
So, for $p$ large, the last congruence 
implies that $aln^l-bmn(m-n)=0$, but that bounds $l$ and therefore $p$, unless
$a=0$. But in this case, the last congruence reads $bmn(m-n) \con 0 (\mod p)$,
which also bounds $p$. As (A) implies that $p$ cannot be bounded, we conclude 
that (A) implies (B) if $m,n$ are integers as in (A).

Suppose now that $m,n\ne 0$ are arbitrary integers and
$a(n\d(m)-m\d(n))+b=0$ for some $a,b$. If $m,n$ are not coprime 
and $m=dm',n=dn',m',n'$ coprime, then
$0=a(n\d(m)-m\d(n))+b= ad^2(n'\d(m')-m'\d(n'))+b$, which reduces (B)
to the case $m,n$ coprime. If $m=m_1^r,n=n_1^r$, then
$0=a(n\d(m)-m\d(n))+b=r(n_1m_1)^{r-1}(n_1\d(m_1)-m_1\d(n_1))+b$,
which reduces (B) to the case $m/n$ is not a perfect power, so 
(A) implies (B) in general. 

If $\sum a_i\d(m_i) = b$ assume, replacing $m_i$ by $-m_i$ and
$a_i$ by $-a_i$ if necessary, that the $m_i$ are all positive. 
Let 
$$m= \prod_{a_i>0} m_i^{a_im_i}, n=\prod_{a_i<0} m_i^{-a_im_i}$$
then
$$\d(m)= \sum_{a_i>0} a_im\d(m_i),\d(n)=\sum_{a_i<0} -a_in\d(m_i).$$
Therefore $n\d(m)-m\d(n)=mn\sum a_i\d(m_i)=mnb$, thus
by (B) we conclude that $m/n=\pm 1$ and therefore the $m_i$'s are
multiplicatively dependent.
So (B) implies (C).

{\bf Remarks:}
(i) Note that to prove (B) for a given pair $m,n$ satisfying the
hypotheses of (A) we only need (A) for the same pair $m,n$.

(ii) Some of the calculations in the proof that (A) implies (B)
generalize some results of Johnson [J].

(iii) The fact that $b\d(2) \ne 0$ for all $b \in \Z, b\ne 0$
is equivalent to there being infinitely many primes $p$ with $2^p \not\con 2
(\mod p^2)$, which is an open problem and indicates that (B) is likely to
be out of reach of present techniques. However, one could get by with something
weaker than (A) when $n=1$, namely that $(m^l-1)/(m-1)$ has a large prime factor
for infinitely many $l$.

(iv) One may conjecture that, under the hypotheses of (C),
that $d(n_1),...,d(n_r)$ are actually algebraically independent over $\Z$.
We can prove that, for $r=1$, this is also implied by (A). In fact, if
$P(\d(m)) =0$, for a polynomial $P$ with integer coefficients, we get as
before $P(-m(m-1)/l) \con 0 (\mod p)$, for $p=(m^l-1)/(m-1)$, prime. 
Again we can use an estimate to get $P(-m(m-1)/l)=0$ and complete the
proof as before. Note that irreducible polynomials in one variable
over $\Z$ and of degree bigger than one have no roots in $R$, by the
Chebotarev density theorem, but some reducible polynomials do,
such as $(x^2-2)(x^2-3)(x^2-6)$, so this extension of the theorem is
non-vacuous.

(v) The ring $R$ has many quotients which are fields of characteristic
zero, the so-called non-principal ultraproducts of the $\F_p$. One can then
ask similar questions for these quotients.


{\bf Acknowledgements:} The author would like to thank A. Buium for
the question that inspired this note. This research was started during
the first Arizona winter school on Arithmetical Algebraic Geometry,
at the Southwestern Center for Arithmetical Algebraic Geometry, and
finished at MSRI. The author would also like to thank the
TARP (grant ARP-006) and the NSA (grant MDA904-97-1-0037) 
for financial support.

\bigskip
\centerline{\bf References.}
\bigskip
\noindent
[B1] A. Buium, {\it Geometry of $p$-jets}, Duke Math. J., {\bf 82} (1996),
349-357.
\medskip
\noindent
[B2] A. Buium, {\it Arithmetic analogues of derivations}, J. Algebra 
{\bf 198} (1997),290--299.
\medskip
\noindent
[I] Y. Ihara, {\it On Fermat quotient and "differentiation of numbers"}
RIMS Kokyuroku {\bf 810} (1992) 324-341, (In Japanese). English translation
by S. Hahn, Univ. of Georgia preprint.
\medskip
\noindent
[J] W. Johnson, {\it On the nonvanishing of Fermat quotients $(\mod p)$}, 
J. Reine Angew. Math. {\bf 292} (1977), 196--200. 
\medskip
\noindent
[Sm] A. L. Smirnov, {\it Hurwitz inequalities for number fields}, St. Petersburg
Math. J. {\bf 4} (1993) 357-375.
\medskip
\noindent
Dept. of Mathematics, Univ. of Texas, Austin, TX 78712, USA
\smallskip
\noindent
e-mail: voloch@math.utexas.edu



\end