ON CONGRUENCE PROPERTIES OF THE PARTITION FUNCTION

Some congruence properties of the partition function are proved.


Introduction and statement of results.
A partition of n is defined to be a nonincreasing set of positive integers whose sum is n.The unrestricted partition function, or p(n), is defined to be the number of partitions of n.
(1.2) If one were to examine in any detail the values of p(n) for any set of n, it would become obvious that these congruences are unexpected and rare.In fact, K. Ono has begun the process of quantifying their rarity (see [4] and [3]).
The first to show that there are infinitely many even and odd values of the partition function was Kolberg [1].However, other conjectures exist regarding the parity of the partition function.
Conjecture 1.2 [5].In every 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.
Theorem 1.3.For any arithmetic progression r (mod t) there are infinitely many integers N ≡ r (mod t) for which p(N) is even.
Theorem 1.4.For any arithmetic progression r (mod t), there are infinitely many integers M ≡ r (mod t) for which p(M)is odd, provided there is one such M.
As is seen in Theorem 1.4, the odd case of Subbarao's conjecture remains open.
The first result we obtain in this paper is the following theorem.This theorem provides us, for the first time, with an explicit infinite set of non-trivial cases of Subbarao's conjecture in the odd case.Theorem 1.5.For all primes l > 3, in any progression kl (mod l n ), where 0 ≤ k ≤ l n−1 − 1 and n is any positive integer, there are infinitely many M ≡ kl (mod l n ) where p(M) is odd.
However, we are interested in more than just the parity of the partition function.As shown in [4], congruences like Ramanujan's are indeed rare, but there are few concrete results that detail precise infinite families of progressions that do not possess similar congruence properties.Our next result, the following theorem, gives us an example of such a family of progressions.
Theorem 1.6.For all primes l > 3 in any progression kl (mod l n−1 ), where 0 ≤ k ≤ l n − 1 and n is any positive integer, there are infinitely many M ≡ kl (mod l n ) where p(M) is not congruent to 0 (mod l).
Example 1.7.In order to illustrate the conclusions of Theorems 1.5 and 1.6, let l = 5 and n = 3.Now let r be any of the numbers 0, 5, 10, 15, 20,...,120.By Theorem 1.5, we know that there are infinitely many non-negative integers n for which p(125n+r ) is odd.By Theorem 1.6, there are infinitely many non-negative integers n for which p(125n + r ) is not divisible by 5.

2.
Proof of Theorems 1.5 and 1.6.Before we begin the discussion of the two main theorems of this paper, we must present the following three theorems that will be used in their proofs.First it is necessary to discuss Hensel's lemma [2].
Hensel's Lemma.Suppose f (x) is a polynomial with integral coefficients.If there is an integer a and a prime p such that f (a) ≡ 0 (mod p j ) and f (a) is not congruent to 0 (mod p), then there is a unique t (mod p) such that f (a+ tp j ) ≡ 0 (mod p j+1 ).
This theorem allows us to establish the solvability of a congruence (mod p j+1 ) by its solvability (mod p j ).
In [4], Ono proves the following two theorems.
Theorem 2.1.Let 0 ≤ r < t be integers for which gcd(24r − 1,t) = 1.If an integer n exists such that n ≡ r (mod t) for which p(n) is odd, then for every integer s coprime to 24t there are infinitely many M ≡ s 2 (r −24 −1 )+24 −1 (mod t) for which p(M) is odd.Theorem 2.2.Let 0 ≤ r < t be integers for which gcd(24r − 1,t) = 1 and let l be an odd prime.If an integer n exists such that n ≡ r mod t for which p(n) is not congruent to 0 (mod l) then for every integer s coprime to 24t there exists infinitely many M ≡ s 2 (r − 24 −1 ) + 24 −1 (mod t) for which p(M) is not congruent to 0 (mod l).
These theorems relate directly to the question of the parity and congruences of the partition function.As explained in the preceding section, it is known that there are infinitely many integers n ≡ r (mod t) where p(n) is even and infinitely many n ≡ r (mod t) for which p(n) is odd, providing that there is at least one such n.Theorem 1.5 allows us to determine some cases where such an odd p(n) exists.
Proof of Theorem 1.5.In Theorem 2.1, first note that p(0) = 1.Let n, r = 0 and let t = l n .Then for any l n , gcd(−1,t) = 1, so the first condition is fulfilled.Now, we know that for every s coprime to 24t, there exists infinitely many M ≡ s 2 (r − 24 −1 ) + 24 −1 (mod t) for which M is odd.Substituting values, we see that there exists infinitely many M ≡ −24 −1 (s 2 − 1) (mod l n ) for s coprime to 24l n .To prove the trivial case of k = 0, take s = 1.The proof for the other cases is more complicated.Our goal is to prove that for s coprime to 24l n , −24 −1 (s 2 − 1) covers the values 1l, 2l,...,(l n−1 − 1)l.
The first step in this process is to eliminate the −24 −1 coefficient.Multiply the congruence by −24.We obtain s 2 − 1 ≡ −24kl (mod l n ).Since −24kl is still a multiple of l, the question is now reduced to whether or not we can find s coprime to 24l n such that s 2 − 1 ≡ kl (mod l n ) for all k and n.
There are two different methods of proving that this congruence is solvable.First, we use a constructive method.Take the congruence s 2 − 1 ≡ kl (mod l), or equivalently, s 2 − 1 − kl ≡ 0 (mod l).The solutions s to this congruence are of the form ml ± 1.
If we input ml + 1 into the congruence (mod l 2 ) we find another solution. (2.1) We can pick m so that ml + 1 is coprime to 24t.In fact, it is easy to see that If we input (ml 2 + (kl/2) + 1) = s into the congruence we obtain Therefore, by letting m = (−k 2 /8) (mod l) we obtain a solution to the congruence.This process can be repeated, and therefore tells us that the congruence is always solvable for each n.Remark 2.3.In an alternate proof, we can use Hensel's lemma to reduce the question of this congruence's solvability to the solvability of s 2 − 1 − kl ≡ 0 mod l.By taking s = ±1, this congruence can be solved in all cases.Note also that for s = ±1, gcd(s, 24l n ) = 1.The derivative of s 2 − 1 − kl is equal to 2s, which is not congruent to 0 (mod l).
Suppose there is a solution to s 2 −1−kl ≡ 0 (mod l n ).As before, since f = 2s which is not congruent to 0 (mod l), Hensel's lemma tells us that there is a set of s that are solutions to the congruence (mod l n+1 ).By induction, this gives us another proof of Theorem 1.5.
A similar method can be used to prove Theorem 1.6.