# Integer as Sum of Polygonal Numbers

## Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then $n$ is:

- $(1): \quad$ Either triangular or the sum of $2$ or $3$ triangular numbers
- $(2): \quad$ Either square or the sum of $2$, $3$ or $4$ square numbers
- $(3): \quad$ Either pentagonal or the sum of $2$, $3$, $4$ or $5$ pentagonal numbers
- and so on.

That is:

- for all $k \ge 3$, $n$ is the sum of no more than $k$ polygonal numbers of order $k$.

## Proof

First some lemmata:

### Lemma 1

Let $n, m \in \N_{>0}$ such that $m \ge 3$.

Let $n < 116 m$.

Then $n$ can be expressed as a sum of at most $m + 2$ polygonal numbers of order $m + 2$.

$\Box$

### Lemma 2

Let $n, m \in \R_{>0}$ such that $\dfrac n m \ge 1$.

Define $I$ to be the open real interval:

- $I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$

Then:

- For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$.

$\Box$

### Lemma 3

Let $n, m, r \in \R_{>0}$.

Suppose $\dfrac n m > 1$.

Let $b \in \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$.

Define:

- $a = 2 \paren {\dfrac {n - b - r} m} + b = \paren {1 - \dfrac 2 m} b + 2 \paren {\dfrac {n - r} m}$

Then $a, b$ satisfy:

- $b^2 < 4 a$
- $3 a < b^2 + 2 b + 4$

$\Box$

First note that Cauchy's Lemma (Number Theory) gives us:

Let $a$ and $b$ be odd positive integers.

Suppose $a$ and $b$ satisfy:

- $b^2 < 4 a$
- $3 a < b^2 + 2 b + 4$

Then there exist non-negative integers $s, t, u, v$ such that:

\(\ds a\) | \(=\) | \(\ds s^2 + t^2 + u^2 + v^2\) | ||||||||||||

\(\ds b\) | \(=\) | \(\ds s + t + u + v\) |

$\Box$

Now we consider the numbered statements as asserted.

For $(1)$, we have Integer is Sum of Three Triangular Numbers.

For $(2)$, we have Lagrange's Four Square Theorem.

For each $m \ge 3$, we prove that $n$ is the sum of at most $m + 2$ polygonal numbers of order $m + 2$.

We choose to consider polygonal numbers of order $m + 2$, rather than $m$-gonal numbers, because the former have a neater closed form:

- $\map P {m + 2, k} = \dfrac m 2 \paren {k^2 - k} + k$

For $n < 116 m$ the result is shown in Lemma 1.

It remains to investigate the integers $n \ge 116 m$.

Define:

- $I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$

By Lemma 2:

- For $\dfrac n m \ge 116$, the length of $I$ is greater than $4$.

Thus $I$ must contain at least $2$ consecutive odd integers.

Let $b_1, b_2$ be those consecutive odd integers.

Because $b_1 + m - 1 = b_2 + m - 3$:

- the set $\set {b + r: b \in \set {b_1, b_2}, r \in \set {0, 1, \dots, m - 2} }$ contains a complete residue system modulo $m$.

Hence one can choose some $b, r$ and write $n \equiv b + r \pmod m$.

Define:

- $a = 2 \paren {\dfrac {n - b - r} m} + b = \paren {1 - \dfrac 2 m} b + 2 \paren {\dfrac {n - r} m}$

Since $m \divides \paren {n - b - r}$ and $b$ is odd:

- $a$ is an odd integer.

Since $b \in I$, from Lemma 3 we have:

- $b^2 < 4 a$
- $3 a < b^2 + 2 b + 4$

Then by Cauchy's Lemma (Number Theory), we can write:

- $a = s^2 + t^2 + u^2 + v^2$
- $b = s + t + u + v$

for nonnegative integers $s, t, u, v$.

Thus:

\(\ds n\) | \(=\) | \(\ds \frac m 2 \paren {a - b} + b + r\) | From $a = 2 \paren {\dfrac {n - b - r} m} + b$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac m 2 \paren {s^2 + t^2 + u^2 + v^2 - s - t - u - v} + s + t + u + v + r\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \paren {\frac m 2 \paren {s^2 - s} + s} + \paren {\frac m 2 \paren {t^2 - t} + t} + \paren {\frac m 2 \paren {u^2 - u} + u} + \paren {\frac m 2 \paren {v^2 - v} + v} + r \times 1\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map P {m + 2, s} + \map P {m + 2, t} + \map P {m + 2, u} + \map P {m + 2, v} + r \map P {m + 2, 1}\) |

Since $r \le m - 2$, the above expression comprises at most $m + 2$ polygonal numbers of order $m + 2$.

$\blacksquare$

## Historical Note

Many of Fermat's theorems were stated, mostly without proof, in the margin of his copy of Bachet's translation of Diophantus's *Arithmetica*.

In $1670$, his son Samuel published an edition of this, complete with Fermat's marginal notes.

### Fermat's Note

As Fermat himself put it:

*Every positive integer is triangular or the sum of $2$ or $3$ triangular numbers; a square or the sum of $2$, $3$ or $4$ squares; a pentagonal number or the sum of $2$, $3$, $4$ or $5$ pentagonal numbers; and so on to infinity, whether it is a question of hexagonal, heptagonal or any polygonal numbers.*

*I cannot give the proof here, for it depends on many abstruse mysteries of numbers; but I intend to devote an entire book to this subject, and to present in this part of number theory astonishing advances beyond previously known boundaries.*

Euler struggled on and off with the proof for squares without success for nearly $40$ years. Lagrange finally managed this in $1772$.

Gauss proved the theorem for triangular numbers in $1796$.

Finally, Cauchy proved the theorem in its completeness in $1815$.

## Sources

- Jan 1987: Melvyn B. Nathanson:
*A Short Proof of Cauchy's Polygonal Number Theorem*(*Proceedings of the American Mathematical Society***Vol. 99**,*no. 1*: pp. 22 – 24)

- 1992: George F. Simmons:
*Calculus Gems*... (previous) ... (next): Chapter $\text {A}.13$: Fermat ($\text {1601}$ – $\text {1665}$)