Legendres Theorem and Quadratic Reciprocity

Jonah Sinick

Swarthmore College, Department of Mathematics & Statistics

Rational Points on Conic

Sections

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

TIFF (LZW) decompressor

are needed to see this picture.

QuickTime and a

TIFF (LZW) decompressor

are needed to see this picture.

QuickTime and a

TIFF (LZW) decompressor

are needed to see this picture.

Diophantus

Legendre

Hilbert

Legendres Theorem

Connection with Quadratic

Reciprocity

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

reciprocity.

References

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

Society

McTutor Mathematical Biography Archive

Acknowledgements

Thanks to Walter Stromquist for his mathematical mentoring

over the years. IHs taught me a lot.