# Chapter 7. "Of the Extraction of the Roots of Polygon Numbers."

### Part I. Section IV. Chapter 7. “Of the Extraction of the Roots of Polygon Numbers.”

656 We have shown, in a preceding chapter (Section III, Chapter 5. “Of Figurate, or Polygonal Numbers.”), how
polygonal numbers are to be found; and what we then called
a **side**, is also called a **root**. If, therefore, we represent the
root by \(x\), we shall find the following expressions for all
polygonal numbers

3-gon | \( \dfrac{x^2+x}{2} \) |

4-gon | \( x^2 \) |

5-gon | \( \dfrac{3x^2-x}{2} \) |

6-gon | \( 2x^2-x \) |

7-gon | \( \dfrac{5x^2-3x}{2} \) |

8-gon | \( 3x^2-2x \) |

9-gon | \( \dfrac{7x^2-5x}{2} \) |

10-gon | \( 4x^2-3x \) |

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

657 We have already shown, that it is easy, by means of these formulae, to find, for any given root, any polygon number required: but when it is required reciprocally to find the side, or the root of a polygon, the number of whose sides is known, the operation is more difficult, and always requires the solution of a quadratic equation; on which account the subject deserves, in this place, to be separately considered. In doing this we shall proceed regularly, beginning with the triangular numbers, and passing from them to those of a greater number of angles.

658 Let therefore 91 be the given triangular number, the side or root of which is required.

If we make this root =\(x\), we must have \(\dfrac{x^2+x}{2}=91\); or \(x^2+x=182\), and \(x^2=-x+182\); consequently,

\[x=-\frac{1}{2}+\surd\left(\frac{1}{4}+182\right) = -\frac{1}{2} + \surd\left(\frac{729}{4}\right) =-\frac{1}{2} + \frac{27}{2}=13;\]from which we conclude, that the triangular root required is 13; for the triangle of 13 is 91.

659 But, in general, let \(a\) be the given triangular number, and let its root be required.

Here if we make it \(=x\), we have \(\dfrac{x^2+x}{2}=a\), or \(x^2+x=2a\); therefore, \(x^2=-x+2a\), and by the rule \(x=-\frac{1}{2} +\surd\left(\frac{1}{4}+2a\right)\), or \(x=\dfrac{-1 + \surd(8a+1)}{2}\).

This result gives the following rule: To find a triangular root, we must multiply the given triangular number by 8, add 1 to the product, extract the root of the sum, subtract 1 from that root, and lastly, divide the remainder by 2.

660 So that all triangular numbers have this property; that if we multiply them by 8, and add unity to the product, the sum is always a square; of which the following small Table furnishes some examples:

Triangles | 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | 45 | 55 | etc. |

8 times + 1 = | 9 | 25 | 49 | 81 | 121 | 169 | 225 | 289 | 361 | 441 | etc. |

If the given number a does not answer this condition, we conclude, that it is not a real triangular number, or that no rational root of it can be assigned.

661 According to this rule, let the triangular root of 210 be required; we shall have \(a = 210\), and \(8a + 1 = 1681\), the square root of which is 41; whence we see, that the number 210 is really triangular, and that its root is \(\dfrac{41-1}{2}=20\). But if 4 were given as the triangular number, and its root were required, we should find it \(=\dfrac{\surd 33}{2}-\dfrac{1}{2}\), and consequently irrational. However, the triangle of this root, \(\dfrac{\surd 33}{2}-\dfrac{1}{2}\), may be found in the following manner:

Since \(x=\dfrac{\surd 33-1}{2}\), we have \(x^2=\dfrac{17-\surd 33}{2}\), and adding \(x=\dfrac{\surd 33-1}{2}\) to it, the sum is \(x^2+x=\frac{16}{2}=8\). Consequently, the triangle, or the triangular number, is \(\dfrac{x^2+x}{2}=4\).

662 The quadrangular numbers being the same as squares, they occasion no difficulty. For, supposing the given quadrangular number to be \(a\), and its required root \(a\);, we shall have \(x^2=a\), and consequently, \(x=\surd a\); so that the square root and the quadrangular root are the same thing.

663 Let us now proceed to pentagonal numbers.

Let 22 be a number of this kind, and \(x\) its root; then, by the third formula, we shall have \(\dfrac{3x^2-x}{2}=22\), or \(3x^2-x=44\), or \(x^2=\frac{1}{3}x+\frac{44}{3}\); from which we obtain, \(x=\frac{1}{6}+\surd\left(\frac{1}{16}+\frac{44}{3}\right)\), or

\[x=\dfrac{1+\surd (529)}{6} = \frac{1}{6}+\frac{23}{6}=4;\]and consequently 4 is the pentagonal root of the number 22.

664 Let the following question be now proposed: the pentagon \(a\) being given, to find its root.

Let this root be \(x\), and we have the equation \(\dfrac{3x^2-x}{2}=a\), or \(3x^2-x=2a\), or \(x^2=\frac{1}{3}x+\frac{2a}{3}\); by means of which we find \(x=\frac{1}{6}+\surd\left(\frac{1}{36}+\frac{2a}{3}\right)\), that is,

\[x=\dfrac{1+\surd(24a+1)}{6}.\]Therefore, when \(a\) is a real pentagon, \(24a+1\) must be a square.

Let 330, for example, be the given pentagon, the root will be

\[x=\dfrac{1+\surd(7921)}{6} = \dfrac{1+89}{6}=15.\]665 Again, let \(a\) be a given hexagonal number, the root of which is required.

If we suppose it \(=x\), we shall have \(2x^2-x=a\), or \(x^2=\frac{1}{2}x+\frac{1}{2}a\); and this gives

\[x=\frac{1}{4}+\surd\left(\frac{1}{16}+\frac{1}{2}a\right) = \dfrac{1+\surd(8a+1)}{4}.\]So that, in order that \(a\) may be really a hexagon, \(8a + 1\) must become a square; whence we see, that all hexagonal numbers are contained in triangular numbers; but it is not the same with the roots.

For example, let the hexagonal number be 1225, its root will be

\[x=\dfrac{1+\surd 9801}{4} = \dfrac{1+99}{4}=25.\]666 Suppose \(a\) an heptagonal number, of which the root is required.

Let this root be \(x\), then we shall have \(\dfrac{5x^2-3x}{2}=a\), or \(x^2=\frac{3}{5}x+\frac{2}{5}a\), which gives

\[x=\frac{3}{10}+\surd\left(\frac{9}{100}+\frac{2}{3}a\right)=\dfrac{3+\surd(40a+9)}{10};\]therefore the heptagonal numbers have this property, that if they be multiplied by 40, and 9 be added to the product, the sum will always be a square.

Let the heptagon, for example, be 2059; its root will be found

\[=x=\dfrac{3+\surd(82369)}{10} = \dfrac{3+287}{10} = 29.\]Let the heptagon, for example, be 2059; its root will be found

\[=x=\dfrac{3+\surd(82369)}{10} = \dfrac{3+287}{10} = 29.\]667 Let us suppose a an octagonal number, of which the root \(x\) is required.

We shall here have \(3x^2-2x=a\), or \(x^2=\frac{2}{2}x+\frac{1}{3}a\), whence results

\[x=\frac{1}{3}+\surd\left(\frac{1}{9}+\frac{1}{3}a\right) = \dfrac{1+\surd(3a+1)}{3}.\]Consequently, all octagonal numbers are such, that if multiplied by 3, and unity be added to the product, the sum is consequently a square.

For example, let 3816 be an octagon; its root will be

\[x=\dfrac{1+\surd 11449}{3} = \dfrac{1+107}{3} = 36.\]668 Lastly, let \(a\) be a given \(n\)-gonal number, the root of which it is required to assign; we shall then, by the last formula, have this equation:

\[\dfrac{(n-2)x^2-(n-4)x}{2} = a,\]or \((n-2)x^2-(n-4)x=2a\); consequently,

\[x^2=\dfrac{(n-4)x}{n-2}+\dfrac{2a}{n-2};\]whence,

\[x=\dfrac{n-4}{2(n-2)} + \surd \left( \dfrac{(n-4)^2}{4(n-2)^2}+\dfrac{2a}{n-2}\right),\]or

\[x=\dfrac{n-4}{2(n-2)} + \surd \left( \dfrac{(n-4)^2}{4(n-2)^2}+\dfrac{8(n-2)a}{4(n-2)^2}\right),\]or

\[x=\dfrac{n-4+\surd(8(n-2)a+(n-4)^2)}{2(n-2)}.\]This formula contains a general rule for finding all the possible polygonal roots of given numbers.

For example, let there be given the 24-gonal number, 3009: since \(a\) is here = 3009 and \(n=24\), we have \(n-2=22\) and \(n-4=20\); wherefore the root

\[x=\dfrac{20+\surd(529584+400)}{44} = \dfrac{20+728}{44} = 17.\]#### 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.