TWO CONGRUENCES

INVOLVING 4–CORES

Michael D. Hirschhorn* and James A. Sellers

School of Mathematics UNSW Sydney 2052 NSW, Australia

and

Department of Science and Mathematics Cedarville College P.O. Box 601 Cedarville, OH 45314

Submitted: February 25, 1995; Accepted: June 30, 1995

γνωσϵι πoλλoυς αϵι ϵμπνoιης

Abstract. The goal of this paper is to prove two new congruences involving 4–cores using elementary techniques; namely, if a_{4}(n) denotes the number of 4–cores of n, then a_{4}(9n+2) ≡ 0 (mod 2) and a_{4}(9n + 8) ≡ 0 (mod 4).

1991 Mathematics Subject Classification. 05A17, 11P83. Key words and phrases. Congruence, partitions, t–cores. *This work was done while the first author was on leave at Université Louis Pasteur, Strasbourg,

France.

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Section 1. Introduction.

Several papers have recently been published concerning t–cores and congruence properties that they satisfy. For examples of these, see Garvan [1], Kolitsch [4], and Garvan, Kim, and Stanton [2]. Most of the congruence results developed in the above works deal with prime t. In contrast, the goal of this paper is to focus on two congruences satisfied by 4–cores.

Before looking at the congruences themselves, a brief word is in order concerning the definition of a t–core. Given the partition π of the integer n, we say that π is a t–core if its Ferrers graph does not contain a hook of length t. See James and Kerber [3] for a fuller discussion of the definition of t–cores.

We will focus all attention in this paper on the case t = 4. The generating function for the number of 4–cores of n, denoted by a_{4}(n), is given by

∑_{n≥0} a_{4}(n) q^{n} = \frac{(q^{4}; q^{4})_{∞}^{4}}{(q; q)_{∞}}

where

(a; q)_{∞} = (1 − a)(1 − aq)(1 − aq^{2})(1 − aq^{3}) …

Section 2. Congruences Modulo 2.

In this section, our goal is to prove the following:

Theorem 1. For all n ≥ 0,

a_{4}(9n + 2) ≡ 0 (mod 2),  (1)
and  a_{4}(9n + 8) ≡ 0 (mod 2).  (2)

Proof. Note the following:

∑_{n≥0} a_{4}(n) q^{n} = \frac{(q^{4}; q^{4})_{∞}^{4}}{(q; q)_{∞}}

= \frac{(q^{4}; q^{4})_{∞}(q^{4}; q^{4})_{∞}^{3}}{(q; q)_{∞}}

= (q; q)_{∞}^{3}(q^{4}; q^{4})_{∞}^{3}.\frac{(q^{4}; q^{4})_{∞}}{(q; q)_{∞}^{4}}

≡ (q; q)_{∞}^{3}(q^{4}; q^{4})_{∞}^{3} (mod 2)

= (∑_{n≥0}(−1)^{n}(2n + 1)q^{\binom{n+1}{2}}) ( ∑_{m≥0}(−1)^{m}(2m + 1)q^{4\binom{m+1}{2}})

≡ ∑_{m,n≥0} q^{\binom{n+1}{2} + 4\binom{m+1}{2}} (mod 2).

Here we have used Jacobi’s famous result:

(q; q)_{∞}^{3} = ∑_{n≥0}(−1)^{n}(2n + 1)q^{\binom{n+1}{2}}

Our theorem is now proven provided we show that all coefficients of the terms of the form q^{9n+2} and q^{9n+8} in the double sum above are divisible by 2.

In order to get a contribution to the term q^{9n+2}, we must have

\binom{n + 1}{2} + 4\binom{m + 1}{2} ≡ 2 (mod 9).

However, note that \binom{n+1}{2} ≡ 0, 1, 3 or 6 (mod 9) for every n ∈ ℤ. Hence, \binom{n+1}{2}+4\binom{m+1}{2} ≡ 0, 1, 3, 4, 5, 6, or 7 (mod 9), but never 2. Thus, there is no contribution to q^{9n+2}. This proves (1). Exactly the same approach proves (2); the fact needed to prove (2) is that 8 is not in the above list. ■

Section 3. A Congruence Modulo 4.

Congruence (1) is “best possible” in the sense that there are some values of a_{4}(9n + 2) which are even but not divisible by 4. The earliest example is a_{4}(2), which equals 2. However, (2) is not best possible. In fact, the goal of this section is to prove the following strengthening of (2):

Theorem 2. For all n ≥ 0,

a_{4}(9n + 8) ≡ 0 (mod 4).  (3)

Proof. We have

∑_{n≥0} a_{4}(n)q^{8n+5} = q^{5} \frac{(q^{32}; q^{32})_{∞}^{4}}{(q^{8}; q^{8})_{∞}}

= q \frac{(q^{16}; q^{16})_{∞}^{2}}{(q^{8}; q^{8})_{∞}} (q^{2} \frac{(q^{32}; q^{32})_{∞}^{2}}{(q^{16}; q^{16})_{∞}} )^{2}

= ∑_{n odd, ; positive} q^{n^{2}} ( ∑_{n odd, ; positive} q^{2n^{2}} )^{2}

= ∑_{k,l,m odd, ; positive} q^{k^{2}+2l^{2}+2m^{2}}

So we see that a_{4}(n) equals the number of solutions of the equation 8n+5 = k^{2} +2l^{2} +2m^{2} with k, l, m odd and positive. This was noted by Ono [5], who used it to show that a_{4}(n) is positive for all n. In particular, a_{4}(9n + 8) is the number of solutions of

72n + 69 = k^{2} + 2l^{2} + 2m^{2}.  (4)

We want to show that the number of solutions of (4) is divisible by 4.

As a quick example, note that if n = 0 in (4), we see that

69 = 1^{2} + 2 × 3^{2} + 2 × 5^{2} = 1^{2} + 2 × 5^{2} + 2 × 3^{2}
= 7^{2} + 2 × 3^{2} + 2 × 1^{2} = 7^{2} + 2 × 1^{2} + 2 × 3^{2}

and the number of solutions is 4.

Now, consider equation (4). Modulo 6 this becomes

k^{2} + 2l^{2} + 2m^{2} ≡ 3 (mod 6).

An odd square is 1 or 3 (mod 6). From the tables

\begin{matrix}k^{2} ≡ 1 (mod 6) : \\ l^{2} : & 1 & 1 & 3 & 3 \\ m^{2} : & 1 & 3 & 1 & 3 \\ k^{2} + 2l^{2} + 2m^{2} : & 5 & 3 & 3 & 1 \end{matrix}  \begin{matrix}k^{2} ≡ 3 (mod 6) : \\ l^{2} : & 1 & 1 & 3 & 3 \\ m^{2} : & 1 & 3 & 1 & 3 \\ k^{2} + 2l^{2} + 2m^{2} : & 1 & 5 & 5 & 3 \end{matrix}

we see that the only solutions of k^{2}+2l^{2}+2m^{2} ≡ 3 (mod 6) are (k^{2}, l^{2}, m^{2}) ≡ (1, 1, 3), (1, 3, 1) or (3, 3, 3) (mod 6), that is, (k, l, m) = (±1, ±1, 3), (±1, 3, ±1), or (3, 3, 3) (mod 6).

However, if (k, l, m) ≡ (3, 3, 3) (mod 6), then (k^{2}, l^{2}, m^{2}) ≡ (9, 9, 9) (mod 72) [since (6k + 3)^{2} = 72\binom{k+1}{2} + 9], and then k^{2} + 2l^{2} + 2m^{2} ≡ 45 ≢ 69 (mod 72).

So, in (4), just one of l and m is 3 (mod 6). Suppose it is m. We have m ≡ 3 (mod 6), m^{2} ≡ 9 (mod 72), and

k^{2} + 2l^{2} = 72n + 69 − 2m^{2} ≡ 51 (mod 72).

We will show that for each m ≡ 3 (mod 6), the number of solutions is even. By the symmetry between l and m in (4), the total number of solutions will be divisible by 4.

We want to show that the number of solutions of

k^{2} + 2l^{2} = 72n + 51  (5)

with k, l odd and positive is even. Allowing k, l to be positive or negative, we want to show that the number of solutions is divisible by 8. For example, if n = 0 in (5) above, we have

51 = (±1)^{2} + 2 × (±5)^{2}
= (±7)^{2} + 2 × (±1)^{2},

yielding 8 solutions.

Each solution of (5) with k, l positive gives rise to a family of four solutions (±k, ±l). Modulo 18, (5) becomes

k^{2} + 2l^{2} ≡ 15 (mod 18).

Note the following table, which gives k^{2} + 2l^{2} (mod 18):

k ±1 ±3 ±5 ±7 9
l
±1  3 11 9 15 11
±3  1 9 7 13 9
±5  15 5 3 9 5
±7  9 17 15 3 17
9  1 9 7 13 9

¿From this table, we see that (k, l) ≡ (±1, ±5), (±7, ±1), or (±5, ±7) (mod 18). Observe that each family includes just one solution with (k, l) ≡ (1, 1) (mod 6). We call that solution the representative of the family. Thus, the representative solution has (k, l) ≡ (1, −5), (7, 1), or (−5, 7) (mod 18). For example, in

51 = (±1)^{2} + 2 × (±5)^{2}
= (±7)^{2} + 2 × (±1)^{2},

the family (k, l) = (±1, ±5) is represented by (1, −5) and the family (k, l) = (±7, ±1) by (7, 1).

We now show that the ‘families of four’ come in pairs. Suppose k^{2} + 2l^{2} = 72n + 51 and that (k, l) ≡ (1, −5), (7, 1) or (−5, 7) (mod 18). Set

k′ = \frac{k − 4l}{3} , l′ = −\frac{2k + l}{3} .

Then, (k′, l′) ≡ (1, 1) (mod 6). That is, k′ and l′ are odd integers and

(k′)^{2} + 2(l′)^{2} = (\frac{k − 4l}{3} )^{2} + 2 (−\frac{2k + l}{3} )^{2} = k^{2} + 2l^{2} = 72n + 51.

Note that (k′, l′) belongs to a different family from (k, l). First, k′ ≠ −k since k′ ≡ k ≡ 1 (mod 6). Also, k′ ≠ k since, if k′ = k, k = −2l, and then

72n + 51 = k^{2} + 2l^{2} = 6l^{2} ≡ 0 (mod 6),

which is false. Moreover, (k′, l′) is the representative of that family.

Therefore, since the transformation

( \begin{matrix}k \\ l \end{matrix} ) → ( \begin{matrix}k′ \\ l′ \end{matrix} ) = ( \begin{matrix}1/3 & −4/3 \\ −2/3 & −1/3 \end{matrix} ) ( \begin{matrix}k \\ l \end{matrix} )

is its own inverse, we know the families come in pairs. This completes the proof of (3). ■

Section 4. Concluding Remarks.

The referee has made the following observation. We showed in the proof of Theorem 1 that a_{4}(n) is congruent modulo 2 to the number of representations of n in the form

n = \binom{l + 1}{2} + 4\binom{m + 1}{2}

or,

8n + 5 = (2l + 1)^{2} + (4m + 2)^{2}

with l, m ≥ 0. It follows that

a_{4}(n) ≡ \frac{r_{2}(8n + 5)}{8} (mod 2)

where r_{2}(n) is the number of representations of n as a sum of two squares. As is well-known, r_{2}(n) is given for n > 0 by

r_{2}(n) = 4(d_{1}(n) − d_{3}(n))

where d_{i}(n) is the number of divisors of n which are congruent to i modulo 4. Thus a_{4}(n) is even precisely when d_{1}(8n + 5) − d_{3}(8n + 5) is divisible by 4, which gives the following

Theorem 3. a_{4}(n) is even if and only if at least one of the following holds:

(α) 8n + 5 has a prime divisor p ≡ 3(mod 4) with ord_{p}(8n + 5) odd,

(β) 8n + 5 has a prime divisor p ≡ 1(mod 4) with ord_{p}(8n + 5) ≡ 3(mod 4), (γ) 8n+5 has two prime divisors p_{1}, p_{2} ≡ 1(mod 4) with ord_{p_{1}}(8n+5) and ord_{p_{2}}(8n+5)

both odd. (Here, ord_{p}(n) is the highest power of p which divides n.)

In particular, if n ≡ 2 or 8 (mod 9) then ord_{3}(8n + 5) = 1 is odd, and a_{4}(n) is even.

We believe that a_{4}(n) ≡ 0 (mod 4) for n in other arithmetic progressions, but we do not yet have proofs of these congruences.

References

[1] F. Garvan, Some Congruences for Partitions that are p–Cores, Proceedings of the London Mathematical Society 66 (1993), 449–478.

[2] F. Garvan, D. Kim, and D. Stanton, Cranks and t– cores, Inventiones Mathematicae 101 (1990), 1–17.

[3] G. James and A. Kerber, The Representation Theory of the Symmetric Group, Addison– Wesley Publishing, 1981.

[4] L. Kolitsch, Generalized Frobenius Partitions, k– cores, k–quotients, and Cranks, Acta Arithmetica LXII (1992), 97–102.

[5] K. Ono, On the positivity of the number of t-core partitions, Acta Arithmetica LXVI (1994), 221–228.