ch1.4.7-extraction-roots-polygonal-numbers

Ch 1.4.7 – Of the Extraction of the Roots of Polygonal Numbers

Summary: Applies the quadratic formula to the inverse problem of polygonal numbers — given a polygonal number, find its root (side) — and derives root-formulas and integrality conditions for each polygonal class up to the general $n$-gon.

Sources: chapter-1.4.7

Last updated: 2026-05-03

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Setup

In ch1.3.5-polygonal-numbers Euler established the formula for the $x$-th $n$-gonal number:

$P(n,x)=\frac{(n-2)x^2-(n-4)x}{2}$

Finding $x$ given $P$ amounts to solving a quadratic equation, which is why the topic reappears here (source: chapter-1.4.7, §657).

Polygonal formulas as quadratics

Polygon	Formula $P=$	Quadratic for $x$
Triangle ($n=3$)	$\frac{x^2+x}{2}$	$x^2+x=2P$
Square ($n=4$)	$x^2$	$x^2=P$
Pentagon ($n=5$)	$\frac{3x^2-x}{2}$	$3x^2-x=2P$
Hexagon ($n=6$)	$2x^2-x$	$2x^2-x=P$
Heptagon ($n=7$)	$\frac{5x^2-3x}{2}$	$5x^2-3x=2P$
Octagon ($n=8$)	$3x^2-2x$	$3x^2-2x=P$

Root formulas and integrality conditions

Applying completing-the-square to each case:

Polygon	Root formula	Integrality condition
Triangle	$x=\frac{-1+\sqrt{8P+1}}{2}$	$8P+1$ must be a perfect square
Square	$x=\sqrt{P}$	$P$ must be a perfect square
Pentagon	$x=\frac{1+\sqrt{24P+1}}{6}$	$24P+1$ must be a perfect square
Hexagon	$x=\frac{1+\sqrt{8P+1}}{4}$	$8P+1$ must be a perfect square
Heptagon	$x=\frac{3+\sqrt{40P+9}}{10}$	$40P+9$ must be a perfect square
Octagon	$x=\frac{1+\sqrt{3P+1}}{3}$	$3P+1$ must be a perfect square

(source: chapter-1.4.7, §659–667)

Key observation: the triangular and hexagonal conditions are identical ($8P+1$ is a square), confirming that every hexagonal number is also a triangular number — though the roots differ.

Triangular numbers: the $8a+1$ property

Multiplying any triangular number by 8 and adding 1 always produces a perfect square (§660):

Triangle	$8P+1$
1	9
3	25
6	49
10	81
15	121
$\dots$	$\dots$

If $8P+1$ is not a perfect square, $P$ is not a triangular number and its triangular root is irrational.

General $n$-gon root (§668)

For a given $n$-gonal number $a$ with root $x$:

$x=\frac{(n-4)+\sqrt{8(n-2)a+(n-4{)}^2}}{2(n-2)}$

Example (24-gonal number 3009, $n=24$):

$x=\frac{20+\sqrt{8\cdot 22\cdot 3009+400}}{44}=\frac{20+728}{44}=17$

Related pages

• polygonal-numbers
• ch1.3.5-polygonal-numbers
• completing-the-square
• quadratic-equations
• ch1.4.5-pure-quadratic-equations
• ch1.4.6-mixt-quadratic-equations