This is an answer to Heinz Lueneburg' question [25 Aug 1999]:
"Does Legendre also prove the existence of primitive roots?"
I'll quote Legendre's "Essai sur la the/orie des nombres" (Paris, an
VI-1798), p. 184-185, where it reads (my translation):
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(art. 132) LEMMA. Let $c$ be a prime, and $P$ a polynomial in $x$,
of degree $m$; there cannot be more than $m$ values of $x$, between
+c/2 and -c/2, to make this polynomial divisible by $c$.
(The proof lies on the factorisation of P = (x-k)P' + Ac, where $k$
is a value for $x$, and goes reasoning in an implicit "descent"
way. It is quite possible that this is also the Lagrange's proof Heinz
told us about.)
(art. 133) THEOREM. (Assuming the same as in art. 132) and $P$ be a
divisor of the binomial x^(c-1) - 1; there are always m values of $x$,
between +c/2 and -c/2, to make this polynomial divisible by $c$.
Proof: Let x^(c-1) - 1 = PQ, ith Q of degree c-1-m. As there are c-1
values of $x$, (that is -1, -2, -3, ..., -(c-1)/2 and +1, +2, +3, ...,
+(c-1)/2), which make the first member divisible by $c$, each of them
must make $P$ or $Q$ divisible by $c$. Among these c-1 values, there
cannot be more than $m$ which make $P$ divisible by $c$, because $P$
is of degree $m$ only; and they cannot be less than $m$ because then
there would be more than c-1-m values of $x$ which make $Q$ divisible
by $c$; what is impossible since $Q$ is only of degree c-1-m. Then
the number of values of $x$ which make $P$ divisible by $c$, and
which are between +c/2 and -c/2, is precisely $m$.
[On the following page -186-, in art. 135, Legendre introduces his famous
"symbol", (N/c) for N^((c-1)/2) being only +1 or -1 (c prime not =2)]
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Isn't it "beau comme l'Antique"?
Udai Venedem
venedem@wanadoo.fr
http://perso.wanadoo.fr/alta.mathematica/
(new catalogue, and thematic pages - one in particular, on number theory -
in French)