Congruent Numbers, the Rank of Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture. Brad Groff

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1 Congruent Numbers, the Rank of Elliptic Curves and the Birch and Swinnerton-Dyer Conjecture Brad Groff

2 Contents 1 Congruent Numbers Basic Facts and Elliptic Curves.1 Statement of Our Theorem Example of a Congruent Number Proof (1) () () (1) () (3) (3) () Rank Examples of Curves with Different Ranks Another Link Tunnell s Theorem Birch and Swinnerton-Dyer Conjecture L-Functions Orders of Zeroes Interpretation Relevant Proven Facts 10 1

3 1 Congruent Numbers... First off, I have to thank Dave Karpuk for his excellent paper on this topic. 1.1 Basic Facts Now, a congruent number is any number that can be expressed as the area of a right triangle with sides of rational length. For example, 6 is congruent, formed from the Pythagorean triple (3,, 5). Obviously, 157 is also a congruent number. I ll give someone $0 if they can figure out the lengths of the sides of such a triangle before the end of class 1. It is also interesting to notice that if any number q is a congruent number, then s q is also a congruent number by multiplying the perpendicular legs each by s. Therefore it can be observed that whether or not an integer is a congruent number depends only upon its residue in the group Q /Q. Every such residue class contains only one square-free number from which all other elements of that class can be derived, so it is a convention to only speak of square-free congruent numbers. Current progress towards a general solution of the problem of determing whether a given integer has been incomplete, though many results have been demonstrated. For example, given a prime p, if p 3mod 8 then p is not congruent, but p is. if p 5mod 8 then p is congruent. if p 7mod 8 then p and p are congruent. So why would one talk about elliptic curves and congruent numbers together?... and Elliptic Curves It turns out that we have a theorem that demonstrates a very strong connection between congruent numbers and a class of elliptic curves of a certain form. 1 so long as that person has not tried to name something after himself this semester

4 .1 Statement of Our Theorem Theorem: For a square-free positive integer n, the following statements are equivalent. 1. n is congruent : n = 1 ab, where a + b = c, a, b, c Q. There exist three rational squares r < s < t such that s r = t s = n 3. There exists a rational point on the elliptic curve C n : y = x 3 n x other than the obvious points in the set {O, ( n, 0), (0, 0), (n, 0)}.. Example of a Congruent Number Before we prove this, I m going to go right ahead and take another page out of Dave s paper and start with an example. Let s try this for n = 6. We have that 6 = (1/) 3, the area of the (3,, 5) triangle. 6 can also be written as / = / which is quite close to a perfect square, 5/ = (5/). The difference of the two is itself a perfect square, (1/). Hence, 6 = (5/) (1/) which gives us our r and our s and implies that t = (7/). We can use this to determine a non-obvious rational point on C 6 : y = x 3 36x. Simply evaluate C 6 at x = s to get y = (35/8) which gives us what we want at P = (5/, 35/8)..3 Proof.3.1 (1) () Given our success using x = s in the example, there is motivation to use the same substitution here. Furthermore, we can let x = c /. Proving that (1) () is now reduced to showing that x n and x + n are both squares. That is to say that s n = r and s + n = t. Then, ( c ) 1 x n = ab = a + b 1 ( ) ab = a + b ab a b = We can use an almost identical treatment for x + n, ( c ) 1 x + n = + ab = a + b + 1 ( ab = a + b + ab a + b = 3 )

5 These are both squares that maintain the inequalities stated..3. () (1) This is just another algebraic manipulation. If we let r = a b, t = a+b ( ) ( ) a b a + b s = + n = n ( ) ( ) a b + n = a ab + b a + b + n = = a + ab + b a ab + b + 8n = a + ab + b 8n = ab n = ab Hence, n is expressible as the area of some triangle. Now we need to show that these a and b satisfy a + b = c for some rational c to show that the triangle involved is in fact a right triangle. ( ) a + b s + n =.3.3 () (3) ( ) ( ) a + b a + b s = n = ab s = a + ab + b ab = a + b s = (s) = c = a + b To show this we must find a point on C n (Q) that is not on y = 0. First, factor C n : y = x(x n)(x + n) Now we will let x = s and y = 0 and derive a contradiction. If y = 0, then either x = 0, x n = 0 or x + n = 0. Clearly, x 0 since this would force us to violate the presupposed inequality r < s. This implies that x n = ( a b ) = 0 or x + n = ( a+b ) = 0. Now if x n = 0 then we have that a = b n is not square-free. This is again a contradiction. Similarly,,

6 if x + n = 0 then both a and b must be non-negative as they both represent lengths. This implies that a = b = 0, but then a = b = n = 0 x = 0 s = 0 which is a contradiction of the inequality r < s. Hence, x = s 0 is a solution for which y 0. Furthermore, y Q because all three factors of y shown are squares of rational numbers. In fact, this point is (s, rst)..3. (3) () Take α, β, γ Q and consider C : y = (x α)(x β)(x γ). Also let P = (x 1, y 1 ) C(Q), P = (x, y ) C(Q). We can easily multiply this out into Weierstrass form and use Mathematica to determine x(p ), x α, x β and x γ. x(p ) = x bx 8cx + b ac x 3 + ax + bx + c We can see, via Mathematica, that (x α), (x β) and (x γ) are all squares of rational numbers. Respectively, x α = x β = x γ = [ ( )] 1 x xα + α(β + γ) βγ (x α)(x β)(x γ) [ ( )] 1 x xβ + α(β γ) + βγ (x α)(x β)(x γ) [ 1 ( )] x xγ + α(γ β) + βγ (x α)(x β)(x γ) This is true because the numerator and the denominator are both rational, (x α)(x β)(x γ) = y = y and x, α, β, γ Q. Now, we can use this form for our curve C n : y = (x 0)(x n)(x + n). This gives us α = 0, β = n, γ = n. Thus, x α = x, x β = x n and x γ = x + n. All of these are then squares, including x. Then, x (x n) = n and (x + n) x = n. This means that these have the same properties as r, s, t. QED. 5

7 . Rank It also turns out that there is a connection between the congruency of a number n and the rank of C n : y = x 3 n x. In particular, n is a congruent number if and only if the rank of C n is positive. In order to demonstrate this, we first need to show that C n (Q) torsion = {O, (0, 0), (n, 0), ( n, 0)} Consider the discriminant, D = a 3 c+a b +18abc b 3 7c. Because, a = c = 0, D = b 3. For C n, D = n 6 and by Nagell-Lutz y n 6. Thus, we take all possible solutions such that y n 3 = p e 1 1 p e...p e k k, where this is the prime factorization of n 3, y {± t p f 1 1 p f... p f k k t = 0 or 1, 0 f i e i } It turns out that none of these values satisfy our equation and so for all torsion points of C n, y = 0 or P = O which yields all of our obvious points and C n (Q) tor =. Now, for some congruent number n, we know there is a point on C n (Q) that is not a torsion point. Thus C n (Q) >. Because C n (Q) = Z Z... Z C n (Q) = Z r C n (Q) we have that r 1. Therefore, n congruent rank of C n > 0. The converse is immediate. Take the rank of C n to be strictly positive. Then C n (Q) > P C n (Q), P / C n (Q) tor By statement (3) of our theorem, this implies that n is congruent and so our bijection is complete..5 Examples of Curves with Different Ranks We can use what we ve discovered to discern rank information about curves of this form. Given the curve C : y = x 3 576x we can determine that C has positive rank. Factoring 576 gives us = ( 6). This is a congruent number in the same residue class as 6 in Q /Q. Thus the rank of this curve is positive. Given C 11 : y = x 3 11x, we can determine that this has rank zero. First, notice that 11 = 11 and 11 3(mod 8). By the result in section 1.1, this is not congruent so C n has rank zero. It follows that C has positive rank by the same proposition. 6

8 .6 Another Link Tunnell s Theorem Another general theorem linking elliptic curves to congruent numbers has been proposed. Tunnell s Theorem 1 Suppose n is an odd and square-free integer. Then L(C n, 1) = 0 if and only if #{x, y, z Z : x + y + 8z = n} = #{x, y, z Z : x + y + 3z = n} Tunnell s Theorem (Corollary) If #{x, y, z Z : x + y + 8z = n} #{x, y, z Z : x + y + 3z = n} then n is not a congruent number. Otherwise, n is a congruent number, assuming the Birch and Swinnerton-Dyer Conjecture. Clearly the correlation between this class of elliptic curves and congruent numbers is very strong, which is not surprising as we have seen similar correlations with Pythagorean triples and curves with sixteen torsion points, etc. This theorem also motivates us to investigate the Birch and Swinnerton-Dyer Conjecture. 3 Birch and Swinnerton-Dyer Conjecture The Birch and Swinnerton-Dyer Conjecture was initially motivated by the idea that as the rank of elliptic curves increases the number of solutions over finite fields should increase. In other words, the function π C (x) = p x N p p with N p = #C(F p ) will increase as the rank of C increases. Because of the way squares are distributed over finite fields, it was suspected that N p p + 1. Birch and Swinnerton-Dyer suspected that, because elliptic curves with high rank have more independent points of infinite order, they would have more points P C(F) than predicted by this estimate. Technically, π C (x) K C (log x) r C 7

9 for some constant K C where r C is the rank of C. The modern version of this conjecture states that for any elliptic curve defined over Q, ord s=1 L(C/Q, s) = r C where r C = rank of C Now, what does all of this mean? 3.1 L-Functions L-functions are generalizations of the Riemann Zeta function and Dirchlet L-series. The Riemann Zeta function is 1 ζ(s) = n s which has some recognizable characteristics, such as at s = 1 there is a pole corresponding to the divergence of the harmonic series. We can generalize this function over our finite field to produce a function connected to our elliptic curve. n=1 ζ(c/f p, s) = 1 a px + px (1 x)(1 px) with x = p s, a p = p + 1 N p a p is essentially the error term for the estimate N p p+1. From here we can get to something which directly pertains to what we are working towards, the Hasse-Weil L-function. L(C/Q, s) = p (1 a p p s + p 1 s ) 1 If we ignore some exceptions at s = 1, L(C/Q, 1) = p p N p L-functions can be expressed as a power series and it is only recently that it was shown that this power series has an analytic continuation for the L-functions of elliptic curves, as a corollary to Wiles proof of Fermat s Last Theorem. Trust me. 8

10 3. Orders of Zeroes The ord function corresponds to the order of a zero of a function. It measures how much zero a function has at a certain point. This is a strong analogue to the ord function defined in our book, which determines the divisibility of a number by a prime p, its p-adic valuation. In this case, ord is measured by the exponent on the term that approaches zero. For example, 3.3 Interpretation ord x=5 (x 5) 10 = 10 ord x=0 (x ) = ord x=k (x [ k) n (x k + ) m = n ord x= k=0 (x )k+] = Now, with these tools in place we can begin to understand the Birch and Swinnerton-Dyer Conjecture. The order of the zero at s = 1 corresponds to the rank of the curve. If we look at the power series of the L-function, this corresponds to the sum s divisibility by (s 1). In other words, the order is the maximum n such that, without introducing (s 1) terms into the denominators of any terms, we can produce (s 1) n a k Therefore if the L-function is not zero when evaluated at s = 1, the elliptic curve has rank 0. This curve only has rational points which are torsion points, and so C(Q) is determined entirely by the Nagell-Lutz Theorem. If the L-function is zero at s = 1 and the first derivative of the L-function is non-zero at s = 1, then we can state that the rank of this elliptic curve is 1 and that there exists a point P of infinite order such that all rational points in C(Q) are multiples of P and some element in C(Q) tor. If the L-function is zero at s = 1 and all of its derivatives up to its n th derivative are zero at s = 1 and the n + 1 derivative is non-zero, then the rank of the curve is n, assuming Birch and Swinnerton-Dyer. Thus there are n independent points P i of infinite order and the union of these points with C(Q) tor forms a basis for C(Q). 9

11 Relevant Proven Facts Not much. Currently it has only been shown that this conjecture is valid for ord s=1 L(C/Q, s) = r C = 0 or 1 Another recent result is that the L-function has an analytic continuation for the rest of C. The pertinent L-function had previously only been shown to be analytic on Re(s) > 3/. Mathematicians had been able to treat it formally at s = 1 but without any real justification. 10

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