Ann. Acad. Rom. Sci.

Ser. Math. Appl.

ISSN 2066-6594

Vol. 18, No. 2/2026

PARTITION FUNCTIONS THAT REPEL

PERFECT-POWERS^∗

Ken Ono^†

Communicated by M. Merca

DOI

10.56082/annalsarscimath.2026.2.31

Abstract

A conjecture by Sun states that the partition function p(n), for

n > 1, is never a perfect power. Recent work by Merca et al. proposes

generalizations of perfect-power repulsion for p(n). In this note, we

prove these generalizations for the functions p_B(n), which count the

number of partitions of n with the largest part ≤ B. If B ≥ 4 and

k ≥ 3, with k - (B −1), then we prove that there are only ﬁnitely many

pairs (n, m) for which |p_B(n)−m^k| ≤ d. These results support Sun and

Merca et al.’s conjectures, as p_B(n) → p(n) when B → +∞. To prove

this, we reduce the problem to Siegel’s Theorem, which guarantees the

ﬁniteness of integral points on curves with genus ≥ 1.

Keywords: partition function, perfect powers.

MSC: 11P82, 05A17, 05A20.

1 Introduction and statement of results

The distribution of perfect powers is a classic topic in number theory. A

milestone is Catalan’s conjecture (1844), proven by Mih˘ailescu in 2002 [5],

which asserts that the Diophantine equation

x^a− y^b= 1

(x, y, a, b > 1)

^∗Accepted for publication on October 23, 2025

^†ken.ono691@virginia.edu, Department of Mathematics, University of Virginia,

Charlottesville, VA 22904, USA and Academy of Romanian Scientists, 050044 Bucharest,

Romania

31

Partition functions that repel perfect-powers

32

only has the single solution (x, y, a, b) = (3, 2, 2, 3), and so the only con-

secutive perfect powers are 8 and 9. Building on this classical narrative,

attention has shifted from perfect powers to their relationship with number

theoretic functions. Sun [6, 7] proposed that partition numbers p(n) where

n > 1, are never perfect-powers. This has led to further developments by

Merca et al. [4].

We recall these conjectures. Throughout, p(n) denotes the number of

(unrestricted) integer partitions of n, where p(0) = 1. The generating func-

tion for this function is given by the Euler product

∞

X

Y

1

P(q) =

p(n)qⁿ=

.

(1)

_n=11 − qⁿ

n=0

An integer is a perfect power if it has the form m^kwith integers m ≥ 2 and

k ≥ 2. For a ﬁxed exponent k ≥ 2, we deﬁne

∆_k(n) := min | p(n) − m^k|

(2)

m≥0

for the distance from p(n) to the nearest kth power. The works of Sun [6,7]

and Merca et al. [4] concern the following conjecture.

Conjecture (Sun, Merca et al.). The partition function p(n) repels perfect-

powers.

1. (No perfect-powers) For every n ≥ 2 and k ≥ 2, p(n) = m^kfor all

integers m ≥ 2.

2. (Perfect-power Repulsion) For k ≥ 2 and d ≥ 0, at most ﬁnitely many

n satisfy ∆_k(n) ≤ d.

Remark 1. We note that these questions are similar in ﬂavor to the work

of Tengely and Ulas [8] on equal values of special partition functions that

have Diophantine interpretations.

We prove exact analogs of the conjecture for the sequence of partition

functions p_B(n) that converges to p(n). Speciﬁcally, for B ≥ 2, we consider

the truncated Euler products

B

X

Y

1

P_B(q) =

p_B(n) qⁿ

=

.

(3)

_m=11 − q^m

n≥0

One sees that p_B(n) is the number of partitions of n whose largest part is at

most B. Moreover, we also have that p_B(n) is the number of partitions of n

K. Ono

33

with at most B parts by considering the usual Ferrers board conjugation. Of

course, we have lim_B→+∞p_B(n) = p(n). In analogy with ∆_k(n), we deﬁne

ꢀ

(B)

k

ꢀ

∆

(n) := min p_B(n) − m

,

(4)

m≥0

which is the distance between p_B(n) and the nearest kth power. We prove

the following theorem that conﬁrms perfect-power repulsion for these func-

tions.

Theorem 1. If B ≥ 4 and k ≥ 3, with k - (B − 1), then the following are

true.

1. There are at most ﬁnitely many n for which p_B(n) = m^k.

2. If d ≥ 0, then we have

#{ n ≥ 1 : ∆⁽_k^B)(n) ≤ d } < +∞.

Example 1. The partition function p₂(n) does not repel perfect-powers. To

see this, we note that

ꢁ

ꢂꢁ

ꢂ

X

1

p₂(n) qⁿ

=

q^a

q^2b

.

(1 − q)(1 − q²)

n≥0

a≥0

b≥0

Taking the Cauchy product gives

ꢁ

ꢂ

X

1

=

#{(a, b) ∈ Z²_≥0: a + 2b = n} qⁿ.

(1 − q)(1 − q²)

n≥0

For a ﬁxed n, writing n = a + 2b with a, b ≥ 0 is the same as choosing

b = 0, 1, . . . , bn/2c, and then setting a = n − 2b. Therefore, we have that

p₂(n) = bn/2c + 1

(n ≥ 0).

In particular, for any integer m ≥ 1 and any k ≥ 2, if we take n := 2(m^k−1),

then

j

k

2(m^k− 1)

p₂(n) =

+ 1 = (m^k− 1) + 1 = m^k.

2

Thus, p₂(n) takes every perfect kth power.

Partition functions that repel perfect-powers

34

Example 2. If B = 3, then we have

X

1

P₃(q) =

=

p₃(n)qⁿ,

(1 − q)(1 − q²)(1 − q³)

n≥0

Along each residue class n ≡ r (mod 6), we have

p₃(6t + r) = Q_r(t) (t ∈ Z_≥0),

where the six quadratic polynomials Q_r∈ Z[t] are

Q₀(t) = 3t²+ 3t + 1,

Q₁(t) = 3t²+ 4t + 1,

Q₂(t) = 3t²+ 5t + 2,

Q₃(t) = 3t²+ 6t + 3,

Q₄(t) = 3t²+ 7t + 4,

Q₅(t) = 3t²+ 8t + 5.

For r ∈ {0, 1, 4, 5}, there are inﬁnitely many t ≥ 0 such that Q_r(t) is a

perfect square. Consequently, p₃(n) is a square for inﬁnitely many n in each

of the residue classes r ∈ {0, 1, 4, 5} (mod 6). Each diophantine equation

Q_r(t) = m²reduces to a Pell-type equation in (x, m) with discriminant 12

after completing the square in t.

(Case r = 0) From 3t²+ 3t + 1 = m²,

(6t + 3)²− 12m²= −3.

This is the negative Pell equation x²− 12m²= −3 (with x = 6t + 3), which

has inﬁnitely many integer solutions. For instance, starting with (x, m) =

√

(3, 1) and multiplying by the fundamental unit 7 + 2 12 yields an inﬁnite

family. The ﬁrst few t are 0, 7, 104, 1455, . . ., giving

p₃(6t) ∈ {1, 13², 181², 2521², . . .}.

(Case r = 1) From 3t²+ 4t + 1 = m²,

(6t + 4)²− 12m²= 4.

The Pell-type equation x²− 12m²= 4 has inﬁnitely many solutions; e.g.

t = 0, 8, 120, 1680, . . . produce p₃(6t + 1) = 1, 15², 209², 2911², . . ..

(Case r = 4) From 3t²+ 7t + 4 = m²,

(6t + 7)²− 12m²= 1,

the (positive) Pell equation with fundamental solution (x, m) = (7, 2), hence

inﬁnitely many solutions. We obtain t = 0, 15, 224, . . . and p₃(6t + 4) =

2², 28², 390², . . ..

K. Ono

35

(Case r = 5) From 3t²+ 8t + 5 = m²,

(6t + 8)²− 12m²= 4,

again yielding inﬁnitely many solutions; e.g. t = 1, 31, 449, . . . and p₃(6t +

5) = 4², 56², 780², . . ..

In each case the corresponding Pell or Pell-type equation has inﬁnitely

√

many solutions because the unit group of Z[ 12] is inﬁnite; iterating by the

√

fundamental unit 7 + 2 12 produces an inﬁnite family. Therefore p₃(n)

assumes square values inﬁnitely often in the residue classes n ≡ 0, 1, 4, 5

(mod 6).

To prove Theorem 1, we use the fact that the partition functions p_B(n)

are quasipolynomial for large n, meaning each behaves like ﬁxed polynomials

along speciﬁc arithmetic progressions. This regularity allows us to reduce

the conjectures to curves with genus ≥ 1, which then allows us to appeal to

Siegel’s Theorem. Summing over these progressions implies Theorem 1 .

2 Nuts and bolts and the proof of Theorem 1

This section gathers the necessary structural inputs for our partition func-

tion perfect-power repulsion theorem and then demonstrates Theorem 1.

Subsection 2.1 records the quasipolynomial description of p_B(n) on residue

classes, its growth, discrete spacing, and includes two Diophantine lemmas

that manage the proximity to perfect kth powers. Subsection 2.2 combines

these tools to prove Theorem 1.

2.1

Nuts and bolts

We start with the quasipolynomial (QP) structure for the counting func-

tion p_B(n) and the resulting growth and spacing on arithmetic progressions.

Throughout, we assume that B ≥ 4 and k ≥ 3 are ﬁxed integers, where

k - (B − 1).

Lemma 1 (QP structure, degree, and spacing). If we let L := lcm(1, 2, . . . , B),

then there are polynomials Q₀, . . . , Q_L−1∈ Q[x] of the same degree B − 1

with positive leading coeﬃcients such that for all n we have

p_B(Ln + r) = Q_r(n)

(0 ≤ r < L).

Partition functions that repel perfect-powers

36

Moreover, for each r, one has

Q_r(n) = α_rn^B−1+ O(n^B−2

)

(α_r> 0),

Q_r(n + 1) − Q_r(n) = (B − 1)α_rn^B−2+ O(n^B−3).

Proof. Consider the generating function (3)

B

X

Y

1

P_B(q) =

p_B(n)qⁿ=

,

_m=11 − q^m

n≥0

and let L := lcm(1, 2, . . . , B). All of the poles of P_B(q) lie at the Lth

roots of unity ζ. The pole at ζ = 1 has order B, and every other pole has

order at most B − 1. Therefore, we can write the (ﬁnite) partial fraction

decomposition

e(ζ)

X X

R_ζ,j(q)

(1 − ζq)^j

P_B(q) =

,

ζ^L=1

j=1

where R_ζ,j(q) are polynomials and e(ζ) ≤ B, with e(1) = B. Extracting

coeﬃcients via

ꢃ

ꢄ

N + j − 1

[q^N](1 − ζq)^−j= ζ ^N

,

j − 1

we obtain

e(ζ)

X X

p_B(N) =

ζ ^NP_ζ,j(N),

ζ^L=1

j=1

where each P_ζ,j∈ Q[N] has degree at most j−1. Fixing a residue class N ≡ r

(mod L), then we have that each ζ ^N= ζ ^ris constant, and we conclude that

p_B(Ln + r) agrees (for all n ≥ 0) with a polynomial Q_r(n) ∈ Q[n] of degree

at most B − 1.

The degree is exactly B −1 with positive leading coeﬃcient, because the

only term contributing in top degree is the pole at ζ = 1 and j = B. Near

q = 1, we have

B

Y

1

∼

· (1 − q)^−B

(q → 1),

Q

B

_m=11 − q^m

_m=1m

N^B−1

ꢅ

ꢆ

N+B−1

so [q^N](1 − q)^−B

=

∼

. Therefore, the leading coeﬃcient

B−1

(B − 1)!

equals

L^B−1

α_r=

> 0.

B! (B − 1)!

K. Ono

37

Finally, if Q_r(n) = α_rn^B−1+ O(n^B−2), then the discrete diﬀerence satisﬁes

Q_r(n + 1) − Q_r(n) = (B − 1)α_rn^B−2+ O(n^B−3),

by a one-term binomial expansion. This proves the claims.

The previous lemma is crucial for the proof of Theorem 1 . The next little

lemma establishes that the polynomials Q_rare not simple shifts of perfect

powers.

Lemma 2. Let Q(x) ∈ Q[x] be a polynomial of degree d ≥ 2 and ﬁx an

integer k ≥ 2. Deﬁne

n

T_k(Q) :=

t ∈ Q : ∃ a ∈ Q^×, R(x) ∈ Q[x] with deg R ≥ 1

o

and Q(x) − t = a R(x)^k

.

Then T_k(Q) is ﬁnite. In fact, we have that

ꢇ

T_k(Q) ⊆

Q(ξ) : ξ ∈ Q and Q⁰(ξ) = 0 ,

and so |T_k(Q)| ≤ d − 1.

Proof. Suppose Q(x) − t = a R(x)^kwith a ∈ Q^×, deg R ≥ 1, and k ≥ 2.

Diﬀerentiating gives

Q⁰(x) = a k R(x)^k−1R⁰(x).

Hence Q(x) − t and Q⁰(x) have the common nonconstant factor R(x)^k−1

,

so Q(x) − t has a multiple root. Thus, there is ξ ∈ Q with Q(ξ) = t and

Q⁰(ξ) = 0 (i.e. t is a critical value of Q). Since Q⁰has degree d − 1, there

are at most d − 1 such values.

Corollary 1. For any ﬁxed X ≥ 0, we have

|T_k(Q) ∩ { t ∈ Q : |t| ≤ X }| < +∞.

To prove Theorem 1, we require the following notions about polynomials

Q ∈ Z[x]. We say Q is generic (relative to k ≥ 2) if Q(x) = a R(x)^kfor every

a ∈ Q^×and R ∈ Q[x]. Otherwise, we say that Q is power-type (relative to k).

The next lemma is a Diophantine inputs. The ﬁrst gives ﬁniteness of perfect

and near-perfect kth powers along any progression whose quasipolynomial

piece is generic. The second lemma gives a pointwise lower bound in the

power-type case.

Partition functions that repel perfect-powers

38

Lemma 3. Let Q(x) ∈ Q[x] have degree d ≥ 2, let k ≥ 3 be an integer,

and ﬁx D ∈ Z_≥0. Assume that for every t ∈ Q with |t| ≤ D the polynomial

Q(x)−t is not of the form a·R(x)^kwith a ∈ Q^×and nonconstant R ∈ Q[x].

Then we have

#{(n, m) ∈ Z_≥0× Z : |Q(n) − m^k| ≤ D} < ∞.

Moreover, except possibly in the pair (d, k) = (2, 2), this set is bounded in

terms of Q, k, D.

1

M

Proof. Write Q =

Q_ewith M ∈ Z

and Q_e∈ Z[x] primitive. Then

≥1

ꢀ

k

|Q(n) − m^k| ≤ D

⇐⇒

Q_e(n) − M m

≤ MD.

ꢀ

Hence, it suﬃces to prove ﬁniteness, for each ﬁxed integer t with |t| ≤ MD,

of the Diophantine equation

M m^k= Q_e(n) − t.

(5)

Fix such a t. Factor M as M = u^kb₀with u ∈ Z

and b₀∈ Z

k-th-power-

≥1

free. Then every integer solution (n, m) to (5) yields an integral solution

(X, Y ) = (n, u m) of the superelliptic equation

b₀Y ^k= Q_e(X) − t.

(6)

Conversely, any integral solution (X, Y ) to (6) with u | Y corresponds to a

solution (n, m) = (X, Y/u) of (5 ). Therefore, the set of solutions to (5) is

ﬁnite if and only if the set of integral points (X, Y ) ∈ Z²on (6) is ﬁnite.

Set d = deg Q_e= deg Q ≥ 2 and, for each ﬁxed t ∈ Z, deﬁne

r_t:= #{ x ∈ Q : Q_e(x) = t }

(counted without multiplicity). Note that the set of t with r_t< d is ﬁnite

(the critical values of Q_e).

Case r_t= 1. If Q_e(X)−t = c (X −a)^dwith c ∈ Z\{0} and d = deg Q_e≥ 2,

then (6) becomes

b₀Y ^k= c (X − a)^d.

Write u = gcd(k, d) and k = uk₁, d = ud₁with gcd(k₁, d₁) = 1. If k | d

and c/b₀is a kth power, then Q_e(X) − t would be a constant times a kth

power, contrary to the hypothesis. Otherwise, by standard ﬁniteness results

for Thue/Thue-Mahler type equations (for example, see Chapter 9 of [1]),

this Diophantine equation has only ﬁnitely many integer solutions.

K. Ono

39

Case r_t≥ 3. Then the smooth projective model of (6) has genus g ≥ 1 (see,

e.g., the standard genus formula for superelliptic curves in §IV of [2 ]). By

Siegel’s theorem (see Chapter 8 of [3]), (6) has only ﬁnitely many integral

points.

Case r_t= 2. Write Q_e(X) − t = c(X − a)²(X − b) with a = b. If k ≥ 3, (6)

is a Thue/Thue-Mahler-type equation, and standard results (for example,

see Chapter 9 of [1]) imply ﬁniteness of integral solutions.

Case (d, k) = (2, 2). This is the classical conic case, which may have in-

ﬁnitely many integral points (Pell-type).

When g ≥ 1, Siegel’s theorem on integral points (for example, see Chap-

ter 8 of [3]) implies that the set of integral solutions (X, Y ) ∈ Z²to (6 )

is ﬁnite. Because there are only ﬁnitely many t with |t| ≤ MD, taking

the union over these t shows that there are only ﬁnitely many (n, m) with

|Q(n) − m^k| ≤ D.

2.2

Proof of Theorem 1

By Lemma 1, there is a period L and polynomials Q₀, . . . , Q_L−1∈ Q[X] of

degree B −1 with positive leading coeﬃcients such that p_B(Ln+r) = Q_r(n)

for 0 ≤ r < L. It suﬃces to prove that for each ﬁxed residue class r and

each ﬁxed d ≥ 0, the set

ꢇ

S_r,d:=

(n, m) ∈ Z_≥0× Z : |Q_r(n) − m^k| ≤ d

is ﬁnite; then a ﬁnite union over r gives (ii), and taking d = 0 gives (i).

Fix r and d ≥ 0. For each integer t with |t| ≤ d consider the Diophantine

equation

Q_r(n) − m^k= t,

n ∈ Z_≥0, m ∈ Z.

(7)

Let

T_r(d) :=

n

t ∈ Z : |t| ≤ d and ∃ a ∈ Q^×, R ∈ Q[X] nonconstant

o

with Q_r(X) − t = a R(X)^k

.

By Lemma 2 and Corollary 1, T_r(d) is ﬁnite.

Case of Non-exceptional shifts. For every t ∈ {−d, −d + 1, . . . , d} \ T_r(d),

the polynomial Q_r(X) − t is not a constant times a kth power. Choose

b₀∈ Z_>0so that Q_e(X) := b₀Q_r(X) ∈ Z[X]. Then any solution of (7) gives

an integer solution of

b₀m^k= Q_e(n) − b₀t.

Partition functions that repel perfect-powers

40

By Lemma 3 (applied with the ﬁxed polynomial Q_e(X)−b₀t), there are only

ﬁnitely many such integer solutions. Hence, #{(n, m) : Q_r(n) − m^k= t} <

∞ for all non-exceptional t with |t| ≤ d.

Case of exceptional shifts. Now ﬁx t ∈ T_r(d). As above, pass to Q_e(X) =

b₀Q_r(X) ∈ Z[X] and consider

b₀m^k= Q_e(n) − b₀t.

Let r_tdenote the number of distinct roots of Q_e(X) − b₀t in Q.

• If r_t≥ 3, then the aﬃne curve b₀Y ^k= Q_e(X) − b₀t has the genus

≥ 1, so by Siegel’s theorem (for example, see Chapter 8 of [3]) there are only

ﬁnitely many integer points.

• If r_t= 2, standard Thue-Mahler ﬁniteness applies (for example, see

Chapter 9 of [1]) so there are only ﬁnitely many integer solutions.

• If r_t= 1, then Q_e(X) − b₀t = c (X − a)^dfor some a, c ∈ Z, c = 0,

with d = deg Q_e= B − 1 ≥ 3. Dividing by b₀gives

c

Q_r(X) − t =

(X − a)^d.

b₀

Since k - d = B − 1 by hypothesis, the Diophantine equation b₀m^k

=

c (n−a)^dis of Thue/Thue-Mahler type with coprime exponents and therefore

has only ﬁnitely many integer solutions (for example, see Chapter 9 of [1]).

Therefore, the case r_t= 1 also yields at most ﬁnitely many solutions.

Combining the three subcases, for each t ∈ T_r(d) the equation (7) has

only ﬁnitely many integer solutions. Therefore, S_r,dis the ﬁnite union, over

the ﬁnitely many t ∈ {−d, . . . , d}, of ﬁnite sets. This proves that S_r,dis

ﬁnite. As noted at the start, summing over r establishes (ii), and taking

d = 0 gives (i). ꢀ

Acknowledgements. The author thanks Mircea Merca, Wei-Lun Tsai,

and Maciej Ulas for their comments on an earlier version of this note. The

author thanks the Thomas Jeﬀerson Fund, the NSF (DMS-2002265 and

DMS-2055118) and the Simons Foundation (SFI-MPS-TSM-00013279) for

their generous support. The author is an Honorary Member of the Academy

of Romanian Scientists.

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