Legendres Theorem and Quadratic Reciprocity Jonah Sinick Swarthmore

Legendres Theorem and Quadratic Reciprocity
Jonah Sinick
Swarthmore College, Department of Mathematics & Statistics
Rational Points on Conic
We call a point (x,y) in the plane R2 a rational point if both of its
coordinates are rational numbers.
There are infinitely many rational points on the curve x2 + y2 -1 = 0.
It is easy to verify that if t is any rational number then (x,y) given by
(x.y) = ((1 t2)/(1 + t2) , 2t/(1 + t2)) (1) is a point on the curve. In fact,
(eqn. 1) gives all the rational points on x2 + y2 - 1 = 0.
The curve x2 + y2 - 2 = 0 has infinitely many rational points as well,
and there is a formula closely analogous to (1) that gives all rational
points on this curve.
However, x2 + y2 - 3 =0 has no rational points whatsoever (the proof
is in the adjacent column). These examples show that for some
integer triples (a,b,c) the curve ax2 + by2 - c = 0 has infinitely many
rational points and and for others the corresponding curve has no
rational points. There cannot be finitely finitely many rational points
on the curve since one can generate infinitely many rational points
from a single rational point.
Thus the question of whether or not there are infinitely many points
on the curve boils down to whether there is a single one. A natural
quest is the search for a finite algorithm for determining whether
there is any rational point at all. Legendres theorem provides such
an algorithm.

Genus 0 Curves
If we consider allow x and y in (2) to take on values in complex
projective space then the solution set to (2) is topologically
equivalent to a sphere. More generally, given a polynomial P(x)
with integer coefficients the set S = {x, y on the complex
projective line such that P(x,y) = 0} is topologically equivalent
to a compact connected orientable surface and by the
classification theorem from topology it must be a the surface of
torus with g holes in it. It turns out that the structure of the
rational points in S is determined by g. When g > 1, by
Faltings theorem there are only finitely many rational points,
but there is no algorithm for obtaining them at present. When g
= 1 the rational points can be analyzed using the theory of
elliptic curves. This poster concerns the nontrivial part of the
case with g = 0.

Projectivizing the Conics
Suppose we have nonzero integers x, y and z such that
x2 + y2 = z2 then dividing through by z2 we obtain the equation
(x/z)2 + (y/z)2 = 1, so taking u = x/z and v = y/z we obtain a
point on u2 + v2 = 1. Moreover, u and v are rational because x, y
and z are whole. For example, the Pythagorean triple (3, 4, 5)
yields the rational point (3/5, 4/5) on the unit circle.
In fact, the problem of finding rational points on a conic
ax2 + by2 = c (2) is equivalent to finding that of finding integer
solutions to ax2 + by2 = cz2 (3) To see this, let T be the set of
rational solutions to (2) and let S be the set of nonzero integer
triples that satisfy (3). Then it can be shown that f: S --- > T
given by (x, y, z) -- > (x/z, y/z) is onto and that the inverse
image of a point in T under f is just a family of elements of the
form (kx, ky, kz) where gcd(x, y, z) = 1 and k ranges over all
nonzero integers. So there is a perfect one to one
correspondence between rational points on the conic and
relatively prime solutions to (3). Thus, determining the
existence of a rational solution to (3) is the same as determining
the existence of nonzero integer solutions to (3)

Nonzero solutions ---> Nonzero solutions (mod m)
Suppose we have (x, y, z) in Z3 such that x2 + y2 = 3z2,,
reducing (mod 3) we see that x2 + y2 = 0 (mod 3), but
squares are 0 or 1 (mod 3) so we must have x = y = 0 (mod
3). But then x2 = y2 = 0 (mod 9) so that 0 = x2 + y2 = 3z2
(mod 9) and z = 0 (mod 3), so that gcd(x, y, z) > = 3. So
there are no relatively prime solutions to x2 + y2 = 3z2 which
means that there are no nonzero solutions whatsoever. More
generally, (3) has no nontrivial solution if it has no
nontrivial solution (mod m).

QuickTime and a
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QuickTime and a
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QuickTime and a
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Legendres Theorem

Connection with Quadratic

Whats special about (3) is that the existence of nontrivial
solutions (mod m) for some m actually implies that there are
nontrivial whole solutions to (3). This is the substance of the
following theorem of Legendre:

Solvability of the congruence in Legendres theorem is equivalent to the
condition that these three congruences are solvable:

Suppose a, b, c are integers that are not all of the same sign,
squarefree, and gcd(a,b,c) = 1. Then the equation

(Here solvability (mod 1) is understood to hold by default.) Thus
Legendres theorem has something to do with quadratic congruences.
Legendre himself attempted to use his theorem to prove quadratic
reciprocity. He succeeded only in obtaining partial results, for example he
showed that if p and q are primes that are 1 and 3 (mod 4) respectively then
if p is nonsquare (mod q) then q is nonsquare (mod p). He did so by
considering the equation x2 + py2 -qz2 = 0, this equation has no nontrivial
whole solutions because it has no nontrivial solutions (mod 4). However,
the equation meets the hypotheses of Legendres theorem, so one of the
quadratic congruences below must be unsolvable:

ax2 + by2 + cz2 = 0 has a nontrivial solution if and only if
ax2 + by2 + cz2 = 0 (mod |abc|) has a nontrivial solution)
Note that there are only finitely many possible nontrivial
solutions to the latter congruence so that this gives us finite
algorithm for determining whether (3) has nontrivial
solutions. While Legendres theorem doesnt cover all
equations of form (3), its easily to see that every equation of
form (3) can be treated by reducing it to one that satisfies
Legendres hypotheses .The necessity of solvability of the
congruence is straightforward and essentially demonstrated in
the box below.
Sketch of proof of sufficiency :Show that ax2 + by2 + cz2
splits into linear factors (mod m) with m = |abc|, so that
finding a nontrivial solution (mod m) is the same as finding a
nontrivial solution to ax + by + cz = 0 (mod m) . Use a
pigeonhole argument to that there is a nontrivial solution to
the latter equation with x, y and z fairly small integers, so that
ax2 + by2 + cz2 = 0(mod m) for x, y and z small in magnitude.
Use the bounds on magnitude together with the modularity
condition to show that
ax2 + by2 + cz2 = 0 or -abc for (x,y,z) nonzero.
In the former case were done, in latter case an ingenious trick
shows that another nonzero triple picked in terms of the first
one does yield zero.

Local Global Principle
Legendres theorem is the first historical example of how
knowledge of the solutions to an equation (mod m) for all m
can yield integer or rational solutions to an equation. These
theme is one that has great prevalence in modern number
theory. The idea is that it is easy to analyze equations (mod
m) or locally and that this information can be patched
together to obtain a global picture of integer solutions. In
fact, the famous Birch Swinnerton-Dyer conjecture is of this
sort - there the idea is that the number of solutions to an
elliptic curve (mod p) (for each prime p) determines the rank
of the group of rational points on the elliptic curve and can
be used to determine the group in entirety.

x2 = -bc (mod |a|).

x2 = -ac (mod |b|),

x2 = -ab (mod |c|)

x2 = pq (mod 1)
x2 = q (mod p)
x2 = -p (mod q)
The first hold automatically, also the third holds because both p and -1 are
nonsquares (mod q), so their product must be a square. So it must be the
second congruence that has no solution which is the same as q being
nonsquare (mod p).
Legendres method of proof cannot be used to prove every instance of
quadratic reciprocity, however,it contains the germ of the important Hilbert
Product formula: (3) fails to be solvable modulo arbitrarily high powers of
p for only a finite number of primes p and moreover, the finite number
must be even. Hilberts Product formula is equivalent to quadratic

Grosswald, Emil. 1984. Topics from the Theory of Numbers (2nd Ed.)
Birkhauser Boston
Weil, Andre. 1984 Number Theory: An approach through history; From
Hammurapi to Legendre. Birkhauser Boston
Kato, Kazuya et. Al. 2000. Fermats Dream. American Mathematical
McTutor Mathematical Biography Archive

Thanks to Walter Stromquist for his mathematical mentoring
over the years. IHs taught me a lot.

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