©Electronic Publishing House q-SERIES, ELLIPTIC CURVES, AND ODD VALUES OF THE PARTITION FUNCTION

Let p(n) be the number of partitions of an integer n. Euler proved the following recurrence for p(n): p(n)=∑k=1∞(−1)k


The partition function.
A partition of a nonnegative integer n is any non-increasing sequence of positive integers whose sum is n.Let p(n) denote the number of partitions of n.Even though Euler's recurrence ( * ) gives a method for computing p(n), there are many open problems and conjectures regarding the overall behavior of the partition function.For instance, the following questions regard the parity of p(n).
Conjecture 1.1 (Parkin-Shanks [5]).The number of n ≤ x for which p(n) is even is ∼ (1/2)x.Conjecture 1.2 (Subbarao [7]).In any arithmetic progression r (mod t), there are infinitely many integers N ≡ r (mod t) for which p(N) is even, and there are infinitely many integers M ≡ r (mod t) for which p(M) is odd.K. Ono [3] has recently proven most of this conjecture.
Euler proved that the generating function for p(n) was given by the infinite product (1.1) Euler also discovered the identity 2. Elliptic curves.An elliptic curve over the rationals is a non-singular curve of the form y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 , ( where the coefficients a i are integers.Any curve of the above form is isomorphic to one, say E, of the form with integers a, b, and c.The discriminant of E, denoted by ∆(E), is given by where N p is the number of points of Ēp rational over GF (p), including the point at infinity.There are similar rules for those p with bad reduction.If p is prime and k ≥ 2, then The L-function is then given by As a consequence of (2.5) and (2.6), we obtain: where a p i ≡ 0 (mod 2) and a q j ≡ 1 (mod 2). (2.9) Then a(n) is odd if and only if every a i ≡ 0 (mod 2) and every b j ≡ 2 (mod 3).
Proof.By hypothesis, every p i and q j are odd primes all with good reduction.Then by (4), we find that for every k ≥ 2, The Taniyama-Shimura-Weil conjecture states that all elliptic curves over the rationals are modular.A curve is modular if its L-function corresponds to the Fourier expansion at infinity of a modular form.Specifically, if E is modular and is a modular form.For a number of explicit examples (see [1]), the form F E (z) is given as a product of Dedekind's η-function defined by For example, take the η-product (2.17) The coefficients of the L-function L(E, s) of the elliptic curve E : y 2 = x 3 + 1 are the same as those in the Fourier expansion of F E (z).
3. q-series results.In this section, we give two theorems which do not depend on elliptic curves.They simply depend on q-series manipulations.

then an odd number of the values
are odd, where a ≥ 0 and b are integers.
Proof.Consider the η-product Factor this as Recall the following identity due to Jacobi.
Using this identity and another well known identity, we obtain and is the generating function for the partition function, we find that (3.9) Using Jacobi's identity, (1.2) and the fact that (3.10) Therefore, we find that Therefore, it is easy to check that (3.12) The theorem now follows immediately.
Example 3.1.Here, we illustrate an example of Theorem 3.2.If m = 1, then n = (6m + 1) 2 = 49.We must find pairs (a, b) with a ≥ 0 and b integers such that These pairs are: Theorem 3.2 tells us that an odd number of the following values are odd: Nine of the eleven are indeed odd.

Group law for elliptic curves.
If E is an elliptic curve, E : y 2 = x 3 + ax 2 + bx + c, the point at infinity is taken to be its identity element O, and P = x 1 ,y 1 and Q = x 2 ,y 2 are points on E, then P + Q := x 3 ,y 3 , where and (3.23) The question of finding points of order two on a curve is the same as that of finding all the points such that P + P = O but P ≠ O.It is easily seen from the above that this is satisfied only when y = 0.

Fundamental theorem. If E is an elliptic curve and p is a prime of good reduction, then
Ēp with the point at infinity is a finite abelian group.Proof.By definition, a(p) = p +1−N p , where N p is the number of rational points of E over GF (p).Since p is an odd prime, we find that a(p) is odd if and only if (3.27) The elliptic curve Ēp is a finite abelian group with N p elements, so Lagrange's theorem states that N p is a multiple of the order of each of the individual points.Thus, asking when N p is odd is the same as asking for which Ēp are there no points of order two.A point of order 2 on an elliptic curve is one whose y-coordinate is zero.Thus, N p and, consequently, a(p) is odd if the equation y 2 ≡ x 3 + ax 2 + bx + c ≡ 0 (mod p) has no solution for which y = 0.
Theorem 3.4.Let p 1 < p 2 < ••• be the primes for which have solutions in GF p i and let q 1 < q 2 < ••• be the primes for which has no solutions in GF q j .Suppose that n > 1 is relatively prime to 2378 and that it has the factorization If every a i ≡ 0 (mod 2) and every b j ≡ 2 (mod 3), then an odd number of the values are odd, where a ≥ 0 and b are integers.

Proof.
In [1], it is proved that if E is the curve then its Hasse-Weil function L(E, s) = ∞ n=1 a(n)/n s has the property that its coefficients a(n) are given by (3.33)However, since 1 − q 11n 2 ≡ 1 − q 22n (mod 2), we find that Therefore, we find by (1.1) that But by Jacobi's identity (3.4) and Euler's identity (1.2), this reduces to Therefore, it turns out that The result now follows immediately from Theorem 3.3 and Proposition 2.1.
Example 3.2.It is easy to show that there is no solution to the equation for the primes 3 and 5.So by (2.6), n = 15 is a suitable choice to illustrate Theorem 3.4.
We must, therefore, find all pairs (a, b) with a ≥ 0 and b ∈ Z such that 14 ≥ have solutions in GF p i and q 1 < q 2 < ••• the primes for which has no solutions in GF q j .Suppose that n > 1 is relatively prime to 14 and that its prime factorization is If every a i ≡ 0 (mod 2) and every b j ≡ 2 (mod 3), then an odd number of the values are odd, where a ≥ 0 and b are integers.
Proof.In [1], it is proved that if E is the curve then its Hasse-Weil function L(E, s) = ∞ n=1 a(n)/n s has the property that its coefficients a(n) are given by (3.45) Using Euler's identity (1.2), Jacobi's identity (3.4), and the fact that (1−q 2n ) ≡ (1−q n ) 2 (mod 2), the theorem follows in a manner similar to that of Theorem 3.4.
Theorem 3.6.Let p 1 < p 2 < ••• be the primes for which have solutions in GF p i and q 1 < q 2 < ••• the primes for which has no solutions in GF q j .Suppose that n > 1 is relatively prime to 10 and that its prime factorization is

.48)
If every a i ≡ 0 (mod 2) and every b j ≡ 2 (mod 3), then an odd number of the values are odd, where a ≥ 0 and b are integers.
Proof.If E is the curve then in [1], it was proved that the coefficients a(n) of its Hasse-Weil function L(E, s) = ∞ n=1 a(n)/n s are given by The proof follows in a manner similar to Theorem 3.4.
Theorem 3.7.Let p 1 < p 2 < ••• be the primes for which have solutions in GF p i and q 1 < q 2 < ••• the primes for which has no solutions in GF q j .Suppose that n > 1 is relatively prime to 6 and that its prime factorization is then in [1], it was proved that the coefficients a(n) of its Hasse-Weil function L(E, s) = ∞ n=1 a(n)/n s are given by η( 2)η(4z)η(6z)η(12z) = q ∞ n=1 1 − q 2n 1 − q 4n 1 − q 6n 1 − q 12n .(3.57) The proof follows in a manner similar to Theorem 3.4.
Theorem 3.8.Let p 1 < p 2 < ••• be the primes for which have solutions in GF p i and q j are the primes for which x 3 − 432 ≡ 0 (mod q j ) (3.59) has no solutions in GF q j .Suppose that n > 1 is relatively prime to 6 and that its prime factorization is Proof.If E is the curve then in [1], it was proved that the coefficients a(n) of its Hasse-Weil function L(E, s) = ∞ n=1 a(n)/n s are given by η 2 (3z)η 2 (9z) = q ∞ n=1 1 − q 3n 2 1 − q 9n 2 . (3.63) The proof follows in a manner similar to Theorem 3.4.
Also, realize that the curves in Theorems 3.4, 3.5, and 3.8 were all changed from the form they are normally shown into the form y 2 = x 3 + ax 2 + bx + c by a simple change of variables to ease the job of finding points of order two.

Theorem 3 . 3 .
Let E be the elliptic curveE : y 2 = x 3 + ax 2 + bx + c, (3.26) and L(E, s) = ∞ n=1 a(n)/n s its Hasse-Weil function.If the odd prime p has good reduction, then a(p) is odd if and only if x 3 +ax 2 +bx +c ≡ 0 (mod p) has no solution.
.3) If p is prime, then let GF (p) denote the finite field with p elements.If p is prime, then Ēp is the reduction of E to GF (p).If the reduction is smooth, then we say E has good reduction at p. Otherwise, E has bad reduction at p.If p ∆(E), then E has good reduction at p.
The Hasse-Weil L-function of E, denoted by L(E, s), is obtained by examining the reductions Ēp .If p is a prime of good reduction, then define the integer a(p) as .54) If every a i ≡ 0 (mod 2) and every b j ≡ 2 (mod 3), then an odd number of the values