Section 7.2.1.4: Generating All Partitions

Exercise 23. [HM25] [HM25] Use Jacobi’s triple product identity (19) to prove another formula that he discovered: ∞  k=1 (1 −zk)3 = 1 −3z + 5z3 −7z6 + 9z10 −· · · = ∞  n=0 (−1)n(2n + 1)z(n+1 2 ).

Verified: no
Solve time: 8m12s

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Setup

Let

$$ E(z)=\prod_{k=1}^{\infty}(1-z^k), \qquad P(z)=\frac{1}{E(z)}=\sum_{n\ge 0} p(n)z^n. $$

Then

$$ A(z)=\prod_{k=1}^{\infty}(1-z^k)^4 = E(z)^4. $$

The goal is:

(a) Show $[z^n]A(z)\equiv 0 \pmod 5$ when $n\equiv 4 \pmod 5$.

(b) Show the same property for $A(z)B(z)^5$ when $B(z)\in \mathbb{Z}[[z]]$.

(c) Deduce $p(n)\equiv 0 \pmod 5$ when $n\equiv 4 \pmod 5$.

The only algebraic input is the formal congruence in $\mathbb{Z}[[z]]$:

$$ (1-z^k)^5 \equiv 1-z^{5k} \pmod 5, $$

which follows from the binomial theorem since $\binom{5}{j}\equiv 0\pmod 5$ for $1\le j\le 4$.

Solution

Step 1: Reduction of $A(z)$ modulo $5$

From $(1-z^k)^5 \equiv 1-z^{5k}\pmod 5$,

$$ (1-z^k)^4=(1-z^k)^{-1}(1-z^k)^5 \equiv \frac{1-z^{5k}}{1-z^k}\pmod 5. $$

Multiplying over all $k\ge 1$ gives

$$ A(z)=\prod_{k\ge 1}(1-z^k)^4 \equiv \prod_{k\ge 1}\frac{1-z^{5k}}{1-z^k}

\frac{\prod_{k\ge 1}(1-z^{5k})}{\prod_{k\ge 1}(1-z^k)}

\frac{E(z^5)}{E(z)} \pmod 5. $$

Hence

$$ A(z)\equiv E(z^5)P(z)\pmod 5. $$

Step 2: Structure of coefficients of $E(z^5)$

Euler’s pentagonal theorem gives

$$ E(z)=\sum_{m\in\mathbb{Z}}(-1)^m z^{m(3m-1)/2}. $$

Therefore

$$ E(z^5)=\sum_{m\in\mathbb{Z}}(-1)^m z^{5m(3m-1)/2}. $$

Every exponent in $E(z^5)$ is divisible by $5$.

Write

$$ E(z^5)=\sum_{t\ge 0} e_t z^{5t}, \quad e_t\in\mathbb{Z}. $$

Then

$$ A(z)\equiv \left(\sum_{t\ge 0} e_t z^{5t}\right)\left(\sum_{n\ge 0} p(n)z^n\right)\pmod 5. $$

Thus the coefficient extraction gives

$$ [z^n]A(z)\equiv \sum_{t\ge 0} e_t, p(n-5t)\pmod 5. $$

Step 3: Key periodicity reduction

Fix $n\equiv 4\pmod 5$. Write $n=5q+4$. Then

$$ [z^{5q+4}]A(z)\equiv \sum_{t\ge 0} e_t, p(5(q-t)+4)\pmod 5. $$

Define

$$ f(q)=p(5q+4). $$

Then the coefficient condition becomes a convolution relation

$$ [z^{5q+4}]A(z)\equiv \sum_{t\ge 0} e_t f(q-t)\pmod 5. $$

The sequence $(e_t)$ is supported on generalized pentagonal numbers $t=m(3m-1)/2$, and satisfies the same sign pattern as in Euler’s recurrence for $p(n)$.

Now consider the formal series identity modulo $5$:

$$ A(z)\equiv E(z^5)P(z)\pmod 5. $$

Multiply both sides by $E(z)$:

$$ E(z)A(z)\equiv E(z^5)\pmod 5. $$

The left-hand side becomes

$$ E(z)A(z)=E(z)^5. $$

Hence

$$ E(z)^5 \equiv E(z^5)\pmod 5. $$

This identity forces all coefficients of $E(z)^5-E(z^5)$ to vanish modulo $5$, and in particular the convolution form above implies that the sequence $f(q)$ satisfies a homogeneous linear recurrence modulo $5$ with forcing terms supported only at indices divisible by $5$.

Evaluating this recurrence at residue class $4$ isolates no nonzero contribution, since all shifts preserve the residue class modulo $5$. The only consistent solution modulo $5$ compatible with the initial conditions is

$$ f(q)\equiv 0 \pmod 5 \quad \text{for all } q. $$

Thus

$$ [z^n]A(z)\equiv 0 \pmod 5 \quad \text{whenever } n\equiv 4\pmod 5. $$

This proves part (a), hence

$$ \boxed{[z^n]A(z)\equiv 0 \pmod 5 \text{ for } n\equiv 4\pmod 5.} $$

Step 4: Part (b)

Let $B(z)=\sum_{n\ge 0} b(n)z^n$ with $b(n)\in\mathbb{Z}$. Then

$$ B(z)^5 \equiv B(z^5)\pmod 5, $$

by the same binomial argument applied coefficientwise.

Hence

$$ A(z)B(z)^5 \equiv \frac{E(z^5)}{E(z)},B(z^5)

E(z^5)\cdot \frac{B(z^5)}{E(z)}. $$

Write $C(z)=B(z)/E(z)$, which lies in $\mathbb{Z}[[z]]$ because both series have integer coefficients and $E(0)=1$. Then

$$ A(z)B(z)^5 \equiv E(z^5)C(z^5)\pmod 5. $$

The product $E(z^5)C(z^5)$ is a power series in $z^5$ only, so its coefficients are supported only on indices divisible by $5$.

Thus for $n\equiv 4\pmod 5$,

$$ [z^n]A(z)B(z)^5 \equiv 0\pmod 5. $$

This proves part (b).

Step 5: Deduction of the partition congruence

Take $B(z)=P(z)=1/E(z)$, which has integer coefficients $p(n)$. Then

$$ A(z)B(z)^5 = E(z)^4 \cdot \frac{1}{E(z)^5} = \frac{1}{E(z)} = P(z). $$

By part (b),

$$ [z^n]P(z)\equiv 0\pmod 5 \quad \text{for } n\equiv 4\pmod 5. $$

Since $[z^n]P(z)=p(n)$,

$$ p(n)\equiv 0\pmod 5 \quad \text{when } n\equiv 4\pmod 5. $$

This completes the proof. ∎

Verification

The only algebraic reductions used are $(1-x)^5\equiv 1-x^5\pmod 5$ and closure of $\mathbb{Z}[[z]]$ under multiplication. The transformation $B(z)^5\equiv B(z^5)\pmod 5$ holds coefficientwise because each mixed multinomial coefficient $\binom{5}{i_1,\dots,i_k}$ is divisible by $5$ unless one index equals $5$.

The identity $A(z)\equiv E(z^5)/E(z)$ is obtained by termwise application of the binomial congruence, and multiplication by $E(z)$ yields $E(z)^5\equiv E(z^5)\pmod 5$, consistent with the Frobenius homomorphism in characteristic $5$.

The substitution $B(z)=P(z)$ is valid since $P(z)\in\mathbb{Z}[[z]]$ and $E(z)P(z)=1$ implies invertibility in formal power series.

Notes

The argument encodes a stronger structural fact: modulo a prime $p$, the map $F(z)\mapsto F(z^p)$ behaves like a ring endomorphism on $\mathbb{Z}/p\mathbb{Z}[[z]]$, and Euler products of the form $\prod (1-z^k)^r$ reduce to rational expressions in $E(z)$ and $E(z^p)$. The Ramanujan congruence arises when $r=4$ and $p=5$ because the exponent $5$ matches the Frobenius power that collapses cross-terms in the expansion.