Newton’s identities and the pentagonal number theorem

Jordan Bell
April 3, 2014

1 Introduction

The pentagonal number theorem is the identity


Euler discovered the pentagonal number theorem in 1740, and finally proved the theorem in 1750 [1]. Euler’s proof is the following. Let


Euler uses the identity n=1(1-an)=1-a1-n=2an(1-a1)(1-an-1) to get n=1(1-xn)=1-x-n=2xn(1-x)(1-xn-1) and so n=1(1-xn)=1-x-x2S1. Euler proves that SN=1-x2N+1-x3N+2SN+1, and then iterates this to obtain the pentagonal number theorem.

In [4], Euler uses Newton’s identities together with the pentagonal number theorem to prove identities for divergent series. Euler says that all the roots of unity of each power will be roots of the equation


and so if we denote the roots by α,β,γ,δ, etc. then


By applying Newton’s identities, Euler gets


and likewise formulas for the sums of the higher powers of the roots. Rademacher [5] clarifies these statements.

Euler uses Newton’s identities in his discovery of the values of ζ(2n). Generally, Euler uses Newton’s identities to relate expressions for the same thing as sums and products, such as the representations of Bessel functions as infinite series and infinite products (cf. [6, pp. 497–503]).

We denote by σ(n) the sum of the positive divisors of n. In 1747, Euler discovered the following recurrence relation for σ(n) [1, pp. 345–356]. Let ωj=j(3j-1)2, the jth pentagonal number. For nm(3m-1)2,


and for n=m(3m-1)2,


In this note we give a proof of Euler’s pentagonal number theorem using Newton’s identities and Euler’s recurrence relation for the sum of divisors function σ. This proof draws attention to the connection between the product (1-xn) and roots of unity.

2 Results

Let k=1n(1+XkT)=k=0nskTk. For d1, let pd=k=1nXkd. Newton’s identities [2, IV.65] are

pd=k=1d-1(-1)k-1skpd-k+(-1)d+1dsd,d1. (1)

Now we shall present a proof of the pentagonal number theorem using Newton’s identities and Euler’s recurrence relation for σ(n). Let ek(j)=e2πijk. Then


Let N(n)=n(n+1)2.

Define sk,n by


On the one hand,


On the other hand,


Hence for kn, sk,n=sk,n+1.

Define sk by


We want to show that sn,n=sn for all n1.

First, 1-t=s0,1-s1,1t and so s1,1=1. By its definition, s1=1. Assume now tha sn,n=sn, and thus that sk,n=sn for all kn.

By Newton’s identities (1),


We note the following.

pd,n = k=1nj=1kek(j)d
= k=1n{k,k|d,0,otherwise.

Hence pd,n=σ(d) if dn. Euler studies exponential sums in [3].

We use the fact that pk,n=σ(k) if kn to get


But furthermore, by assumption sk,n+1=sk for kn, so


Either n+1m(3m-1)2 or n+1=m(3m-1)2. In the first case, the recurrence relation for σ(n+1) is


This implies that sn+1,n+1=0. But sn+1=0 also, so sn+1,n+1=sn+1.

In the second case, the recurrence relation for σ(n+1) is


This implies that (-1)m-1(n+1)=(-1)n+2(n+1)sn+1,n+1, and hence (-1)n+1sn+1,n+1=(-1)m. But (-1)n+1sn+1=(-1)m, so sn+1,n+1=sn+1.

This completes the proof by induction that sn,n=sn for all n1.


  • [1] J. Bell (2010) A summary of Euler’s work on the pentagonal number theorem. Arch. Hist. Exact Sci. 64 (3), pp. 301–373. Cited by: §1, §1.
  • [2] N. Bourbaki (1981) Éléments de mathématique. Lecture Notes in Mathematics, Vol. 864, Masson, Paris. Note: Algèbre. Chapitres 4 à 7. [Algebra. Chapters 4–7] External Links: ISBN 2-225-68574-6, MathReview (Robert Gilmer) Cited by: §2.
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  • [4] L. Euler (1780) De mirabilibus proprietatibus numerorum pentagonalium. Acta Academiae Scientiarum Imperialis Petropolitanae (I), pp. 56–75. Note: E542, Opera omnia I.3, pp. 480–496 Cited by: §1.
  • [5] H. Rademacher (1969) Comments on Euler’s “De mirabilibus proprietatibus numerorum pentagonalium”. In Number Theory and Analysis (Papers in Honor of Edmund Landau), pp. 257–268. External Links: MathReview (F. P. Cass) Cited by: §1.
  • [6] G. N. Watson (1944) A treatise on the theory of Bessel functions. second edition, Cambridge University Press. Cited by: §1.