# Chapter 5. "Of Figurate, or Polygonal Numbers."

### Part I. Section III. Chapter 5. “Of Figurate, or Polygonal Numbers.”

425 The summation of arithmetical progressions, which
begin by 1, and the difference of which is 1, 2, 3, or any
other integer, leads to the theory of *polygonal numbers*,
which are formed by adding together the terms of any such
progression.

426 Suppose the difference to be 1; then, since the first term is 1 also, we shall have the arithmetical progression,

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, etc.

and if in this progression we take the sum of one, of two, of three, etc. terms, the following series of numbers will arise:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, etc.

for 1 = 1, 1 + 2 = 3, 1 + 2 + 3 = 6, 1 + 2 + 3 + 4 = 10, etc.

Which numbers are called **triangular**, or trigonal numbers, because we may always
arrange as many points in the form of a triangle as they contain units, thus:

427 In all these triangles, we see how many points
each side contains. In the first triangle there is only one
point; in the second there are two; in the third there are
three; in the fourth there are four, etc.: so that the triangular numbers,
or the number of points, which is simply
called the triangle, are arranged according to the number of
points which the side contains, which number is called the
**side**; that is, the third triangular number, or the third
triangle, is that whose side has three points; the fourth,
that whose side has four; and so on; which may be represented thus:

Side

\[\begin{array}{c} \bullet \end{array}\]Triangle

\[\begin{array}{c} \bullet \end{array}\]Side

\[\begin{array}{ccc} \bullet&&\bullet\\ \end{array}\]Triangle

\[\begin{array}{ccc} \bullet&&\bullet\\ &\bullet& \end{array}\]Side

\[\begin{array}{ccccc} \bullet&&\bullet&&\bullet \end{array}\]Triangle

\[\begin{array}{ccccc} \bullet&&\bullet&&\bullet\\ &\bullet&&\bullet&\\ &&\bullet&& \end{array}\]Side

\[\begin{array}{ccccccc} \bullet&&\bullet&&\bullet&&\bullet \end{array}\]Triangle

\[\begin{array}{ccccccc} \bullet&&\bullet&&\bullet&&\bullet\\ &\bullet&&\bullet&&\bullet&\\ &&\bullet&&\bullet&&\\ &&&\bullet&&& \end{array}\]Side

\[\begin{array}{ccccccccc} \bullet&&\bullet&&\bullet&&\bullet&&\bullet \end{array}\]Triangle

\[\begin{array}{ccccccccc} \bullet&&\bullet&&\bullet&&\bullet&&\bullet\\ &\bullet&&\bullet&&\bullet&&\bullet&\\ &&\bullet&&\bullet&&\bullet&&\\ &&&\bullet&&\bullet&&&\\ &&&&\bullet&&&& \end{array}\]Side

\[\begin{array}{ccccccccccc} \bullet&&\bullet&&\bullet&&\bullet&&\bullet&&\bullet \end{array}\]Triangle

\[\begin{array}{ccccccccccc} \bullet&&\bullet&&\bullet&&\bullet&&\bullet&&\bullet\\ &\bullet&&\bullet&&\bullet&&\bullet&&\bullet&\\ &&\bullet&&\bullet&&\bullet&&\bullet&&\\ &&&\bullet&&\bullet&&\bullet&&&\\ &&&&\bullet&&\bullet&&&&\\ &&&&&\bullet&&&&& \end{array}\]428 A question therefore presents itself here, whicli is, how to determine the triangle when the side is given? and, after what has been said, this may be easily resolved.

For if the side be \(n\), the triangle will be 1 + 2 + 3 + ⋯ + \(n\). Now, the sum of this progression is \(\dfrac{n^2+n}{2}\); consequently the value of the triangle is \(\dfrac{n^2+n}{2}\).

Thus

\(n=1\) | triangle is | 1, |

\(n=2\) | triangle is | 3, |

\(n=3\) | triangle is | 6, |

\(n=4\) | triangle is | 10, |

and so on: and when \(n = 100\), the triangle will be 5050.

429 This formula \(\dfrac{n^2+n}{2}\) is called the general formula of triangular numbers; because by it we find the triangular number, or the triangle, which answers to any side indicated by \(n\).

This may be transformed into \(\dfrac{n(n+1)}{2}\); which serves also to facilitate the calculation; since one of the two numbers \(n\), or \(n+1\), must always be an even number, and consequently divisible by 2.

So, if \(n = 12\), the triangle is \(\dfrac{12\cdot 13}{2}=6\cdot 13=78\); and if \(n=15\), the triangle is \(\dfrac{15\cdot 16}{2}=15\cdot 8=120\), etc.

430 Let us now suppose the difference to be 2, and we sliall have the following arithmetical progression:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, etc.

the sums of which, taking successively one, two, three, four terms, etc. form the following series:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, etc.

the terms of which are called quadrangular numbers, or
**squares**; since they represent the squares of the natural
numbers, as we have already seen; and this denomination
is the more suitable from this circumstance, that we can
always form a square with the number of points which those
terms indicate, thus:

431 We see here, that the side of any square contains precisely the number of points which the square root indicates. Thus, for example, the side of the square 16 consists of 4 points; that of the square 25 consists of 5 points; and, in general, if the side be \(n\), that is, if the number of the terms of the progression, 1, 3, 5, 7, etc. which we have taken, be expressed by \(n\), the square, or the quadrangular number, will be equal to the sum of those terms; that is to \(n^2\), as we have already seen, Art. 422; but it is unnecessary to extend our consideration of square numbers any farther, having already treated of them at length.

432 If now we call the difference 3, and take the sums
in the same manner as before, we obtain numbers which are
called pentagons, or **pentagonal numbers**, though they cannot be so well represented by points

Indices | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | etc. |

Arithmetic Progression | 1, | 4, | 7, | 10, | 13, | 16, | 19, | 22, | 25, | etc. |

Pentagonal Numbers | 1, | 5, | 12, | 22, | 35, | 51, | 70, | 92, | 117, | etc. |

the indices showing the side of each pentagon.

433 It follows from this, that if we make the side \(n\), the pentagonal number will be \(\dfrac{3n^2-n}{2}=\dfrac{n(3n-1)}{2}\).

Let, for example, \(n = 7\), the pentagon will be 70; and if the pentagon, whose side is 100, be required, we make \(n = 100\), and obtain 14950 for the number sought.

434 If we suppose the difference to be 4, we arrive at
**hexagonal numbers**, as we see by the following progressions:

Indices | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | etc. |

Arithmetic Progression | 1, | 5, | 9, | 13, | 17, | 21, | 25, | 29, | 33, | etc. |

Hexagonal Numbers | 1, | 6, | 15, | 28, | 45, | 66, | 91, | 120, | 153, | etc. |

where the indices still show the side of each hexagon.

435 So that when the side is \(n\), the hexagonal number is \(2n^2 - n = n(2n-1)\); and we have farther to remark, that all the hexagonal numbers are also triangular; since, if we take of these last the first, the third, the fifth, etc. we have precisely the series of hexagons.

436 In the same manner, we may find the numbers which are heptagonal, octagonal, etc. It will be sufficient therefore to exhibit the following Table of formulae for all numbers that are comprehended under the general name of polygonal numbers.

Supposing the side to be represented by \(n\), we have

Triangle | \(\dfrac{n^2+n}{2}=\dfrac{n(n+1)}{2}\) |

Square | \(\dfrac{2n^2+0n}{2}=n^2\) |

Pentagon | \(\dfrac{3n^2-n}{2}=\dfrac{n(3n-1)}{2}\) |

Hexagon | \(\dfrac{4n^2-2n}{2}=2n^2-n=n(2n-1)\) |

Heptagon | \(\dfrac{5n^2-3n}{2}=\dfrac{n(5n-3)}{2}\) |

Octagon | \(\dfrac{6n^2-4n}{2}=3n^2-2n=n(3n-2)\) |

Nonagon | \(\dfrac{7n^2-5n}{2}=\dfrac{n(7n-5)}{2}\) |

Decagon | \(\dfrac{8n^2-6n}{2}=4n^2-3n=n(4n-3)\) |

11-gon | \(\dfrac{9n^2-7n}{2}=\dfrac{n(9n-7)}{2}\) |

12-gon | \(\dfrac{10n^2-8n}{2}=5n^2-4n=n(5n-4)\) |

20-gon | \(\dfrac{18n^2-16n}{2}=9n^2-8n=n(9n-8)\) |

25-gon | \(\dfrac{23n^2-21n}{2}=\dfrac{n(23n-21)}{2}\) |

\(m\)-gon | \(\dfrac{(m-2)n^2-(m-4)n}{2}\) |

437 So that the side being \(n\), the \(m\)-gonal number will be represented by \(\dfrac{(m-2)n^2-(m-4)n}{2}\); whence we may deduce all the possible polygonal numbers which have the side \(n\). Thus, for example, if the bigonal numbers were required, we should have \(m = 2\), and consequently the number sought \(= n\); that is to say, the bigonal numbers are the natural numbers, 1, 2, 3, etc.

If we make \(m = 3\), we have \(\dfrac{n^2+n}{2}\) for the triangular number required.

If we make \(m = 4\), we have the square number \(n^2\), etc.

438 To illustrate this rule by examples, suppose that the 25-gonal number, whose side is 36, were required; we look first in the Table for the 25-gonal number, whose side is \(n\), and it is round to be \(\dfrac{23n^2-21n}{2}\). Then making \(n = 36\), we find 14526 for the number sought.

439 *Question.* A person bought a house, and he is
asked how much he paid for it. He answers, that the 365-gonal number of
side 12 is the number of crowns which it cost him.

In order to find this number, we make \(m=365\), and \(n = 12\); and substituting these values in the general formula, we find for the price of the house 23970 crowns.

#### Editions

- Leonhard Euler.
*Elements of Algebra*. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822. - Leonhard Euler.
*Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange.*Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.