%% alf@mpce.mq.edu.au

%% Copyright 1993
%% Alf van der Poorten
%% ceNTRe for Number Theory Research
%% Macquarie University
%% NSW 2109 AUSTRALIA


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\hskip3cm{\it Sophie Germain was a French mathematician, 
a contemporary of
Cauchy and Legendre, with whom she corresponded. Her theorem, brought by
Legendre to the attention of the 
illustrious members of the Institut de
France, was greeted with great admiration.}\smallskip
\rightline{P. Ribenboim\hskip1cm}\bigskip
\def\norm{\roman N}

{\bf Abel's formul\ae.}\ Suppose $x^p+y^p+z^p=0$. Then
$$\gcd(x+y,\frac{x^p+y^p}{x+y})=1 \quad\hbox{ or\ }\quad p\,,$$ and if
$p\div x^p+y^p$, then $p\div x+y$ and $p\ediv \dfrac{x^p+y^p}{x+y}$.

To see this, just notice that
$$\multline \frac{x^p+y^p}{x+y}=\frac{(x+y-y)^p+y^p}{x+y}\equiv
py^{p-1}-\tfrac12p(p-1)y^{p-2}(x+y) \mod{(x+y)^2}\\
\quad\hbox{ and\ }\quad \equiv (x+y)^{p-1} \mod p\,.\endmultline$$ The
basic underlying fact is that for any $a$, $b$
$$(a+b)^p=a^p+b^p\mod p\,;$$ indeed, this characterises primes $p$.
But more than that, there is Fermat's Theorem (often referred to as
his {\it little\/} theorem) according to which $a^p\equiv a\mod p$
for any rational integer
$a$, so for integers
$a_1$,
$a_2$, $\dots$,
$a_n$ we have the congenial result that
$$(a_1+a_2+\cdots +a_n)^p\equiv a_1+a_2+\cdots +a_n\mod p\,.$$ In any
case, by the factorisations $-z^p=(x+y)\dfrac{x^p+y^p}{x+y}$,
$\ldots$, we have
$$\alignat3 x+y&=c^p&\qquad\dfrac{x^p+y^p}{x+y}&=\gamma^p&\quad\hbox{ if\ }\quad
p&\ndiv z\\ y+z&=a^p&\qquad\dfrac{y^p+z^p}{y+z}&=\alpha^p&\qquad
p&\ndiv x\\ z+x&=b^p&\qquad\dfrac{z^p+x^p}{z+x}&=\beta^p&\qquad
p&\ndiv y\endalignat$$ So, in the {\it first case\/} of Fermat's Last
Theorem, when $p\ndiv xyz$, we have, say, $2x=-a^p+b^p+c^p$.

Incidentally, if $p\div z$, then $x+y=p^{p-1}c^p$ and
$\dfrac{x^p+y^p}{x+y}=\gamma^p$ with $p\ndiv\gamma$. 

Now Sophie Germain argues as follows: If $x^p+y^p+z^p=0$ then 
$x^p+y^p+z^p\equiv0\mod q$, for every $q$. Suppose now that $q=2p+1$
happens to be prime. Then $q\ndiv y$, say, entails $y^p\equiv\pm1\mod
q$, so necessarily
$q\div xyz$, say $q\div x$. Hence $-a^p+b^p+c^p\equiv0\mod q$, so
$q\div abc$; in fact one sees that $q\div a$. Thus $y\equiv -z$, so
$\alpha^p\equiv py^{p-1}\mod q$. Also, of course, $\gamma^p\equiv
y^{p-1}\mod q$. Thus $p$ is a $p$-th power $\mod q$, which entails
$p\equiv\pm1\mod q$, whence we have a contradiction.

Using these ideas Legendre generalised the result to $q=2kp+1$  prime
for several further values of $k$, enabling all cases $p\lt100$ to be
dealt with. In 1823 this was a remarkable advance. 

No-one doubts that there are infinitely many primes $p$ so that also
$2kp+1$ is prime for some suitable small $k$, but such results remain
inaccessible. Nonetheless, fairly recently Adleman and Heath-Brown {\it
inter alia\/}
used a generalisation of Sophie Germain's result, and work of Fouvry, to
prove that
the first case of the FLT held for infinitely many prime exponents.

Sophie Germain employed her own version of the formulas I mention.
Suppose
$0\lt x\lt y\lt z$. Abel considered the case of Fermat's equation
$x^n+y^n=z^n$ when one of $x$,
$y$ or $z$ is  prime.  Abel showed that $x$ must be the prime, that
$z=y+1$, that the exponent $n$ must be a prime $p$, and that
$p\div y(y+1)$. However, his proof that the equation
$$x^p+y^p=(y+1)^p$$ has no nontrivial solutions was shown to be
faulty by Markoff some seventy years later. Thus that Abel's equation
has no solutions remained unproved until several weeks ago. For all
we knew,
$z-y=1$ was possible in the FLT. On the other hand, some 20 years ago,
Kustaa Inkeri and I used Baker's method to prove that
$y-x$ must be large relative to the exponent if Fermat's equation
was to have a solution. Related methods had been used in preceding
years by Inkeri {\it et al\/} to show that in any putative solution
to Fermat's equation each of
$x$,
$y$ and $z$ would have to be very large indeed.



Kummer's results dealt with both the first and second cases, but were
incongenial for extended computation. To the contrary, the results of
Mirimanoff and Wieferich at the turn of the century, and a little
later, of F\"urtwangler, seemed almost to dispose of the first case. They
showed that if there is a first case solution for exponent $p$ then
$$p^2\div 2^{p-1}-1 \quad\hbox{ and\ }\quad p^2\div 3^{p-1}-1\,.$$ These
results relied on a complicated analysis of the conditions first
proved by Kummer. At first no examples of
$p$ satisfying the first condition were known. Then in 1913 Meissner 
noticed
$2^{1092}\equiv1\mod{1093^2}$ and in 1922 Beeger found that $3511^2\div
2^{3510}-1$. Computer checks of some twenty years ago showed that
there are no other cases for the base $2$ with exponent less than
$3\times10^9$. For the base $3$ one has $11^2\div 3^{10}-1$; the next
case is
$3^{1006002}-1\equiv0\mod{1006003^2}$.

Further such criteria for additional bases were derived subsequently.
It is surely most improbable that the two criteria mentioned are ever
satisfied simultaneously. Whatever, we have no understanding of the
{\it Fermat quotients\/} $q_p(a)=(a^p-a)/p$ at all. Their
divisibility by $p$ appears to us as no more than a statistical
accident.\footnote"$^*$"{Apropos Meissner, Landau's {\it
Zahlentheorie\/} points out that with $p=1093$   and all congruences
mod
$1093^2$, clearly $3^7=2187=2p+1$ so $3^{14}\equiv4p+1$. Just so,
$2^{14}=16384=15p-11$, whence
$2^{28}\equiv-330p+121$. Thus
$$3^2\cdot2^{28}\equiv-2970p+1089\equiv-2969p-4\equiv310p-4\,;\quad
3^2\cdot2^{28}\cdot7\equiv2170p-28\equiv-16p-28\,.$$  So
$3^2\cdot2^{26}\cdot7\equiv-4p-7$ and
$3^{14}\cdot2^{182}\cdot7^7\equiv-(4p+7)^7\equiv-7\cdot4p\cdot7^6-7^7$.
Thus
$3^{14}\cdot2^{182}$ is just $-4p-1\equiv-3^{14}$. Hence
$2^{182}\equiv-1$ and Meissner's observation follows.\newline But, is
$3^{1092}\equiv1$ as well? Clearly no, because for any $p$, both
$x^{p-1}\equiv y^{p-1}\equiv1\pmod{p^2}$ as well as $x+y=mp$ with
$p\ndiv m$, is impossible. \newline Now take $x=3^7$, $y=-1$ to see
that
$3^{1092}\not\equiv1\pmod{1093^2}$.}
\bigskip

\centerline{\bf Appendices.}\smallskip

{\bf Fermat's Theorem.}\ Since the object of these notes is to chat
about Fermat's Last Theorem, the least I can do is to say a few words
about a result that properly bears his name. Throughout $p$ is a
rational prime.

The claim $n^p\equiv n\pmod p$ is obvious by induction since
$(n+1)^p\equiv n^p+1\pmod p$.

A rather more to the point proof is to notice that $1$, $2$,
$\dots$, $p-1$ are all the different remainders mod $p$ that are
relatively prime to $p$. But if $a$ is not zero mod $p$ then the
numbers
$a$, $2a$,
$\dots$, $(p-1)a$ are different and nonzero mod $p$. Hence mod $p$
they must be just a permutation of $1$, $2$,
$\dots$, $p-1$. Thus $(p-1)!\equiv (p-1)!a^{p-1}$ and Fermat's
Theorem is evident.

It follows, incidentally, that there is an $a'\in\{1,2,\ldots,p-1\}$
so that
$aa'\equiv1\pmod p$. Of course one writes $a'=a^{-1}$, and notes that
$a^{-1}\equiv a^{p-2}\pmod p$.

It is also fun to remark that the \poly\ $X^p-X$ has distinct zeros
mod
$p$ since it and its derivative $pX^{p-1}-1\equiv-1\mod p$ are
trivially relatively prime. Hence, plainly,
$$X^p-X\equiv X(X-1)(X-2)\cdots\(X-(p-1)\)\mod p$$ and {\it inter
alia\/} --- amongst other things, we have Wilson's Theorem
$$(p-1)!\equiv-1\pmod p\,.$$

It is of course a corollary to Fermat's Theorem that if a prime $p$
divides the
$n$-th Fermat number $F_n=2^{2^n}+1$ then $p=k\cdot2^{n+1}+1$ for some
integer
$k$. Thus, as did Euler, but unlike Fermat --- who must have made an
arithmetical blunder, we can readily find a divisor of
$F_5$.

{\bf Bernoulli numbers.}\ Even engineers know about the power series
$$\sin z=z-\frac1{3!}z^3+\frac1{5!}z^5-\cdots \quad\hbox{ and\ }\quad
\cos z=1-\frac1{2!}z^2+\frac1{4!}z^4-\cdots\,,$$ but what is the
series expansion for $\tan z$?

The trick is first to notice that $f(z)=z/(e^z-1)+\frac12z$ is an even
function, that is, $f(-z)=f(z)$, and then to write --- noting
$B_3=B_5=B_7=\dots=0$,
$$\frac z{e^z-1}=\sum_{k=0}^{\infty}
\frac1{k!}B_kz^k=1-\tfrac12z+\frac1{2!}B_2z^2+\frac1{4!}B_4z^4+
\frac1{6!}B_6z^6+\cdots\,.$$ Clearly the Bernoulli numbers $B_k$ can
be computed by the inductive definition
$$\binom{k+1}1B_k+\binom{k+1}2B_{k-1}+\cdots+\binom{k+1}kB_1+B_0=0\,$$
with  $B_0=1$. So $B_1=-\frac12$, $B_2=\frac16$, $B_4=-\frac1{30}$,
$\ldots\,$. Obviously the
$B_k$ all are rational numbers. However, their numerators soon grow
at a furious rate: indeed, because the cited series evidently has
radius of convergence
$2\pi$, we must have $|B_{2k}|\sim (2k)!/(2\pi)^{2k}$.

What this is all about is the calculus of finite differences, now
unsung, but still taught in applied maths courses when I was littler.
Here the difference operator $\Delta:f(x)\mapsto f(x+1)-f(x)$ and its
inverse
$\Delta^{-1}$, the indefinite sum, replace the derivative
$D:f(x)\mapsto f'(x)$ and its inverse $D^{-1}$, the indefinite
integral. In this context one defines the Bernoulli \poly s $B_n(x)$
by
$$D(\Delta^{-1}x^n)=B_n(x)\,,\quad n=0,1,2,\ldots\;.$$ But on
expanding $f(x+1)$ as a Taylor series about $x$, we see that
$$\Delta f(x)=\sum_{n=1}^{\infty}\frac{f^{(n)}(x)}{n!}=
\sum_{n=1}^{\infty}\frac{D^n}{n!}f(x)=(e^D-1)f(x)\,,$$ so
$$D\Delta^{-1}=\frac{D}{e^D-1} \quad\hbox{ and\ }\quad D\Delta^{-1}
x^n=\sum_{m=0}^{\infty}\frac{B_m}{m!}D^mx^n=\sum_{m=0}^{\infty}
\binom nmB_mx^{n-m}\,,$$ and in particular $B_n=B_n(0)$. Of course
$\Delta^{-1}x^n$ `means' $1^n+2^n+\cdots+(x-1)^n$ because
$$\Delta\(\sum_{m=1}^{x-1}m^n\)=x^n \quad\hbox{ whence\ }\quad
\Delta^{-1}x^n=
\sum_{m=1}^{x-1}m^n=\frac1{n+1}\(B_{n+1}(x)-B_{n+1}(1)\)\,.$$ Shorn
of its error term and other frills, the now fairly evident formula
$$f(x)=
\int_x^{x+1}f(t)\,dt+\sum_{n=1}^\infty\frac{B_n}{n!}\Delta
D^{n-1}f(x)\,,$$ is the useful Euler-Maclaurin formula, allowing one
to compare sums with integrals.\footnote""{Most of this section is
just things I knew, but in the late-seventies I was helped by access
to a draft of Paulo Ribenboim, {\it 13 Lectures on Fermat's Last
Theorem\/}, Springer-Verlag 1980.}

 I'm going to religiously confine each of the Notes to just four
pages so there's just no room for $\tan z$; but surely
$iz\coth iz=z\cot z$ and $\tan z=\cot z-2\cot 2z$ gives enough hint.
Worse though, there is then no space to discuss Euler's evaluation of
the special values of the Riemann $\zeta$-function, that is, of the
sums 
$$\zeta(2k)=\sum_{n=1}^\infty\frac1{n^{2k}}=
1+\frac1{2^{2k}}+\frac1{3^{2k}}+
\frac1{4^{2k}}+\cdots\;=
(-1)^{k-1}\frac{(2\pi)^{2k}}{2(2k)!}B_{2k}\,.$$
\end