THREE CLASSICAL RESULTS ON

REPRESENTATIONS OF A NUMBER

Michael D. Hirschhorn

Introduction

Three classical results concern the number of representations of the positive integer n in the form x^{2} + 3y^{2} with x, y ∈ ℤ, the form (x^{2} + x)/2 + 3(y^{2} + y)/2 with x, y ∈ ℤ^{+} and the form x^{2} + xy + y^{2} with x, y ∈ ℤ. Indeed, if s(n), t(n) and u(n) respectively denote the three numbers, then

(1)  s(n) = 2(d_{1,3}(n) − d_{2,3}(n)) + 4(d_{4,12}(n) − d_{8,12}(n)),

(2)  t(n) = d_{1,3}(2n + 1) − d_{2,3}(2n + 1)

and

(3)  u(n) = 6(d_{1,3}(n) − d_{2,3}(n)).

where d_{r,m}(n) is the number of divisors d of n with d ≡ r (mod m).

(1) is equivalent to the q-series identity

∑_{m,n∈ℤ} q^{m^{2}+3n^{2}} = 1 + 2 ∑_{n≥0} (\frac{q^{3n+1}}{1 − q^{3n+1}} − \frac{q^{3n+2}}{1 − q^{3n+2}} )

+ 4 ∑_{n≥0} (\frac{q^{12n+4}}{1 − q^{12n+4}} − \frac{q^{12n+8}}{1 − q^{12n+8}} ) (4)

or to

(5) ∑_{m,n∈ℤ} q^{m^{2}+3n^{2}} = 1 + 2 ∑_{n≥0} (\frac{q^{3n+1}}{1 − (−1)^{n}q^{3n+1}} − \frac{q^{3n+2}}{1 + (−1)^{n}q^{3n+2}} ) ,

Typeset by AMS-TEX

(2) is equivalent to the q-series identity

(6) ∑_{m,n∈ℤ^{+}} q^{(m^{2}+m)/2+3(n^{2}+n)/2} = ∑_{n≥0} (\frac{q^{3n}}{1 − q^{6n+1}} − \frac{q^{3n+2}}{1 − q^{6n+5}} )

and (3) is equivalent to the q-series identity

(7)  ∑_{m,n∈ℤ} q^{m^{2}+mn+n^{2}} = 1 + 6 ∑_{n≥0} (\frac{q^{3n+1}}{1 − q^{3n+1}} − \frac{q^{3n+2}}{1 − q^{3n+2}} ) .

Both (5) and (6) appear in Ramanujan’s second Notebook [12, p.239], and Berndt [2, pp.223-224 and p.116] shows how they follow from Ramanujan’s 1ψ_{1} summation [12, p.196], [2, p.32]. (7) appears in Berndt and Rankin [5, p.196] and a proof is given by Berndt [3]. The reader is referred also to [6], [7], [9] and [4] for related developments and generalisations, and to [1] and [10] for applications in statistical mechanics.

It seems that Dirichlet (1840) may have known (1), since he gives [8, p.463] the corresponding results for the forms x^{2} + y^{2} and x^{2} + 2y^{2}, and continues “And so on in similar fashion.” (“Et ainsi de suite.”) However, Lorenz (1871) [11, p.420] states both (4) and (1), and in reference to (1) says (my translation) “From this equation one can deduce a theorem which must be considered new in the theory of numbers because it cannot immediately be deduced from known theorems:

If a number N contains prime factors p_{1}, p_{2}, ⋯ of the form 3m + 1 with exponents a_{1}, a_{2}, ⋯ and if the prime factors of the form 3m + 2 appear to nothing but even powers, the number of solutions of the equation m^{2} + 3n^{2} = N is given by

ρ_{N} = 2(a_{1} + 1)(a_{2} + 1) …

if N is odd and by

ρ_{N} = 6(a_{1} + 1)(a_{2} + 1) …

if N is even. If, on the contrary, N contains a prime factor of the form 3m + 2 to an odd power, one has ρ_{N} = 0.” Lorenz also [11, p.424] states (7), and a proof is provided by his reviewer/ translator Valentiner [11, p.430]. So perhaps credit rests with Lorenz.

We shall give proofs of (4), (6) and (7) which demonstrate that all three results are intimately related.

2. Proof of the result involving s(n)

Let a(q) denote the left hand side of (7). Then

(8)  a(q) + 2a(q^{4}) = 3 ∑_{k,l∈ℤ} q^{k^{2}+3l^{2}}.

For,

a(q) = ∑_{{m odd \atop n even}} q^{m^{2}+mn+n^{2}} + ∑_{{m odd \atop n odd}} q^{m^{2}+mn+n^{2}} + ∑_{{m even \atop n odd}} q^{m^{2}+mn+n^{2}}

+ ∑_{m even ; n even} q^{m^{2}+mn+n^{2}}.

In the first sum, let k = m + \frac{n}{2}, l = \frac{n}{2} (and conversely, m = k − l, n = 2l),

in the second sum, k = \frac{m − n}{2}, l = \frac{m + n}{2} (conversely m = k+l, n = l−k),

in the third sum, k = \frac{m}{2} + n, l = \frac{m}{2} (conversely m = 2l, n = k − l)

and in the fourth sum, k = \frac{m − n}{2}, l = \frac{m + n}{2} (conversely m = k + l, n = l − k), and we find

(9)  a(q) = 3 ∑_{k≢l (mod 2)} q^{k^{2}+3l^{2}} + ∑_{k≡l (mod 2)} q^{k^{2}+3l^{2}}.

Also, (10)

a(q^{4}) = ∑_{m,n∈ℤ} q^{4m^{2}+4mn+4n^{2}} = ∑_{m,n even} q^{m^{2}+mn+n^{2}} = ∑_{k≡l (mod 2)} q^{k^{2}+3l^{2}},

as with the fourth sum above. (8) follows from (9) and (10). (4) follows from (7) and (8).

3. Proof of the result involving t(n)

We can write (4)

∑_{k,l∈ℤ} q^{k^{2}+3l^{2}} = 1 + 2 ∑_{n≥0} (\frac{q^{6n+1}}{1 − q^{12n+2}} − \frac{q^{6n+5}}{1 − q^{12n+10}} )

+ 6 ∑_{n≥0} (\frac{q^{12n+4}}{1 − q^{12n+4}} − \frac{q^{12n+8}}{1 − q^{12n+8}} ) .

If we extract the even powers of q we obtain

(11)  ∑_{k≡l (mod 2)} q^{k^{2}+3l^{2}} = 1 + 6 ∑_{n≥0} (\frac{q^{12n+4}}{1 − q^{12n+4}} − \frac{q^{12n+8}}{1 − q^{12n+8}} ) .

(Note, incidentally, that (7) follows from (10) and (11), and that (11) follows from (7) and (10)!) From (11) we deduce (12)

∑_{k,l∈ℤ} q^{k^{2}+3l^{2}} + 4q ∑_{k,l∈ℤ^{+}} q^{k^{2}+k+3l^{2}+3l} = 1 + 6 ∑_{n≥0} (\frac{q^{3n+1}}{1 − q^{3n+1}} − \frac{q^{3n+2}}{1 − q^{3n+2}} ) .

If we subtract (4) from (12) we find

4q ∑_{k,l∈ℤ^{+}} q^{k^{2}+k+3l^{2}+3l} = 4 ∑_{n≥0} (\frac{q^{3n+1}}{1 − q^{3n+1}} − \frac{q^{3n+2}}{1 − q^{3n+2}} ) (13)

− 4 ∑_{n≥0} (\frac{q^{12n+4}}{1 − q^{12n+4}} − \frac{q^{12n+8}}{1 − q^{12n+8}} )

= 4 ∑_{n≥0} (\frac{q^{6n+1}}{1 − q^{12n+2}} − \frac{q^{6n+5}}{1 − q^{12n+10}} ) .

Finally, if we divide (13) by 4q and replace q^{2} by q we obtain (6).

4. Proof of the result involving u(n)

We begin by showing that

(14)  ∑_{m,n∈ℤ} ω^{m−n}q^{m^{2}+mn+n^{2}} = \frac{(q)_{∞}^{3}}{(q^{3})_{∞}}

where ω^{3} = 1, ω ≠ 1. Let CT_{a} {∑_{−∞}^{∞} a^{n}f_{n}(q)} denote f_{0}(q), the “Constant Term” of the Laurent series in a. Then

∑_{m,n∈ℤ}ω^{m−n}q^{m^{2}+mn+n^{2}} = ∑_{m+n+p=0} ω^{m−n}q^{(m^{2}+n^{2}+p^{2})/2}

= CT_{a} {∑_{−∞}^{∞} a^{m}ω^{m}q^{m^{2}/2} ∑_{−∞}^{∞} a^{n}ω^{−n}q^{n^{2}/2} ∑_{−∞}^{∞} a^{p}q^{p^{2}/2}}

= CT_{a} \{∏_{n≥1}(1 + aωq^{n− \frac{1}{2}} )(1 + a^{−1}ω^{−1}q^{n− \frac{1}{2}} )(1 − q^{n}).

. ∏_{n≥1}(1 + aω^{−1}q^{n− \frac{1}{2}} )(1 + a^{−1}ωq^{n− \frac{1}{2}} )(1 − q^{n}).

. ∏_{n≥1}(1 + aq^{n− \frac{1}{2}} )(1 + a^{−1}q^{n− \frac{1}{2}} )(1 − q^{n})\}

= CT_{a} \{∏_{n≥1}(1 + a^{3}q^{3n− \frac{3}{2}} )(1 + a^{−3}q^{3n− \frac{3}{2}} )(1 − q^{n})^{3}\}

= \frac{(q)_{∞}^{3}}{(q^{3})_{∞}} .CT_{a} \{∏_{n≥1}(1 + a^{3}q^{3n− \frac{3}{2}} )(1 + a^{−3}q^{3n− \frac{3}{2}} )(1 − q^{3n})\}

= \frac{(q)_{∞}^{3}}{(q^{3})_{∞}} .CT_{a} \{∑_{−∞}^{∞} a^{3n}q^{3n^{2}/2}\}

= \frac{(q)_{∞}^{3}}{(q^{3})_{∞}} ,

as claimed. Now the left hand side of (14) can be written

∑_{m−n≡0 (mod 3)} q^{m^{2}+mn+n^{2}} + ω ∑_{m−n≡1 (mod 3)} q^{m^{2}+mn+n^{2}}

+ ω^{−1} ∑_{m−n≡−1 (mod 3)} q^{m^{2}+mn+n^{2}}.

In the first sum, let k = \frac{m − n}{3}, l = \frac{m + 2n}{3} (conversely m = 2k + l, n = l − k), in the second sum, k = \frac{m − n − 1}{3}, l = \frac{m + 2n − 1}{3} (conversely m = 2k + l + 1, n = l − k) and in the third sum, k = \frac{n − m − 1}{3}, l = \frac{n + 2m − 1}{3} (m = l−k, n = 2k+l+1), and the left hand side of (14) is seen to be

∑_{k,l∈ℤ} q^{3k^{2}+3kl+3l^{2}} + ω ∑_{k,l∈ℤ} q^{3k^{2}+3kl+3l^{2}+3k+3l+1}

+ ω^{−1} ∑_{k,l∈ℤ} q^{3k^{2}+3kl+3l^{3}+3k+3l+1}

= a(q^{3}) − qc(q^{3}),

where

c(q) = ∑_{m,n∈ℤ} q^{m^{2}+mn+n^{2}+m+n}.

Thus (14) becomes

(15)  a(q^{3}) − qc(q^{3}) = \frac{(q)_{∞}^{3}}{(q^{3})_{∞}} .

Now, it is a celebrated identity of Jacobi that

(16)  (q)_{∞}^{3} = ∑_{n≥0}(−1)^{n}(2n + 1)q^{(n^{2}+n)/2}.

We split this sum according to the residue modulo 3 of n. For n ≡ 0 (mod 3), we write 3n (n ≥ 0), for n ≡ 1 (mod 3), we write 3n + 1 (n ≥ 0), and for n ≡ −1 (mod 3) we write −3n − 1 (n ≤ −1), and the right hand side of (16) becomes

∑_{n≥0}(−1)^{n}(6n + 1)q^{(9n^{2}+3n)/2} − ∑_{n≥0}(−1)^{n}(6n + 3)q^{(9n^{2}+9n+2)/2}

− ∑_{n≤−1}(−1)^{n}(−6n − 1)q^{(9n^{2}+3n)/2}

= ∑_{−∞}^{∞}(−1)^{n}(6n + 1)q^{(9n^{2}+3n)/2} − 3q(q^{9})_{∞}^{3}.

So (15) becomes

(17) a(q^{3}) − qc(q^{3}) = \frac{1}{(q^{3})_{∞}} \{∑_{−∞}^{∞}(−1)^{n}(6n + 1)q^{3(3n^{2}+n)/2} − 3q(q^{9})_{∞}^{3}\} .

It follows that

(18)  a(q) = \frac{1}{(q)_{∞}} ∑_{−∞}^{∞}(−1)^{n}(6n + 1)q^{(3n^{2}+n)/2}

and

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

Now (18) becomes

a(q) = \frac{1}{(q)_{∞}} [\frac{d}{da} ∑_{−∞}^{∞}(−1)^{n}a^{6n+1}q^{(3n^{2}+n)/2}]_{a=1}

= \frac{1}{(q)_{∞}} [\frac{d}{da} \{a ∏_{n≥1}(1 − a^{6}q^{3n−1})(1 − a^{−6}q^{3n−2})(1 − q^{3n})\}]_{a=1}

= \frac{1}{(q)_{∞}} [∏_{n≥1}(1 − a^{6}q^{3n−1})(1 − a^{−6}q^{3n−2})(1 − q^{3n})×

× \{1 + 6 ∑_{n≥1} (\frac{a^{−6}q^{3n−2}}{1 − a^{−6}q^{3n−2}} − \frac{a^{6}q^{3n−1}}{1 − a^{6}q^{3n−1}} )\}]_{a=1}

= \frac{1}{(q)_{∞}} (q)_{∞} \{1 + 6 ∑_{n≥1} (\frac{q^{3n−2}}{1 − q^{3n−2}} − \frac{q^{3n−1}}{1 − q^{3n−1}} )\}

= 1 + 6 ∑_{n≥0} (\frac{q^{3n+1}}{1 − q^{3n+1}} − \frac{q^{3n+2}}{1 − q^{3n+2}} )

which is (7).

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