# Chapter 6. "Of the Squares of Compound Quantities."

### Part I. Section II. Chapter 6. “Of the Squares of Compound Quantities.”

306 When it is required to find the square of a compound quantity, we have only to multiply it by itself, and the product will be the square required.

For example, the square of \(a + b\) is found in the following manner:

\[\begin{array}{rrr} a&+b&\\ a&+b&\\ \hline a^2&+ab&\\ &+ab&+b^2\\ \hline a^2&+2ab&+b^2 \end{array}\]307 So that when the root consists of two terms added together, as \(a + b\), the square comprehends, 1st, the squares of each term, namely, \(a^2\) and \(b^2\); and 2nd, twice the product of the two terms, namely, \(2ab\): so that the sum \(a^2+2ab+b^2\) is the square of \(a+b\). Let, for example, \(a=10\), and \(b=3\); that is to say, let it be required to find the square of 10 + 3, or 13, and we shall have 100 + 60 + 9, or 169.

308 We may easily find, by means of this formula, the squares of numbers, however great, if we divide them into two parts. Thus, for example, the square of 57, if we consider that this number is the same as 50 + 7, will be found = 2500 + 700 + 49 = 3249.

309 Hence it is evident, that the square of \(a + 1\) will be \(a^2+2a+1\): and since the square of \(a\) is \(a^2\), we find the square of \(a+1\) by adding to that square \(2a+1\); and it must be observed, that this \(2a+1\) is the sum of the two roots \(a\) and \(a+1\).

Thus, as the square of 10 is 100, that of 11 will be 100 + 21: the square of 57 being 3249, that of 58 is 3249 + 115 = 3364; the square of 59 = 3364 + 117 = 3481; the square of 60 = 3481 + 119 = 3600, etc.

310 The square of a compound quantity, as \(a + b\), is represented in this manner \((a+b)^2\). We have therefore

\[(a+b)^2=a^2+2ab+b^2,\]whence we deduce the following equations:

\[\begin{gather} (a+1)^2=a^2+2a+1;\\ (a+2)^2=a^2+4a+4;\\ (a+3)^2=a^2+6a+9;\\ (a+4)^2=a^2+8a+16;\\ \textrm{etc.} \end{gather}\]311 If the root be \(a - b\), the square of it is \(a^2 - 2ab + b^2\), which contains also the squares of the two terms, but in such a manner, that we must take from their sum twice the product of those two terms. Let, for example, \(a = 10\), and \(b = - 1\), then the square of 9 will be found equal to 100 - 20 + 1 = 81.

312 Since we have the equation

\[(a-b)^2=a^2-2ab+b^2,\]we shall have \((a - 1)^2 = a^2 - 2a + 1\). The square of \(a - 1\) is found, therefore, by subtracting from \(a^2\) the sum of the two roots \(a\) and \(a - 1\), namely, \(2a - 1\). Thus, for example, if \(a = 50\), we have \(a^2 = 2500\), and \(2a - 1 = 99\); therefore \(49^2 = 2500 - 99 = 2401\).

313 What we have said here may be also confirmed and illustrated by fractions; for if we take as the root ⅗ + ⅖ = 1, the square will be, ⁹⁄₂₅ + ⁴⁄₂₅ + ¹²⁄₂₅ = 1.

Farther, the square of ½ - ⅓ = ⅙ will be ¼ - ⅓ + ⅑ = ¹⁄₃₆.

314 When the root consists of a greater number of terms, the method of determining the square is the same. Let us find, for example, the square of \(a + b + c\):

\[\begin{array}{rrrrrr} a&+b&+c&&&\\ a&+b&+c&&&\\ \hline a^2&ab&+ac&&&\\ &&ab&+b^2&+bc&\\ &&&+ac&+bc&+c^2\\ \hline a^2&+2ab&+2ac&+b^2&+2bc&+c^2\\ \end{array}\]We see that it contains, first, the square of each term of the root, and beside that, the double products of those terms multiplied two by two.

315 To illustrate this by an example, let us divide the number 256 into three parts, 200+50+6; its square will then be composed of the following parts:

\[\begin{array}{rcl} 200^2&=&40000\\ 50^2&=&2500\\ 6^2&=&36\\ 2(50 \cdot 200) &=& 20000\\ 2(6 \cdot 200) &=& 2400\\ 2(6 \cdot 50) &=& 600\\ \hline 65536&=&256\cdot 256 \end{array}\]316 When some terms of the root are negative, the square is still found by the same rule; only we must be careful what signs we prefix to the double products. Thus,

\[(a-b-c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc;\]and if we represent the number 256 by 300-40-4, we shall have,

\[\begin{array}{r} \textrm{Positive Parts.}\\ 300^2=90000\\ 400^2=16000\\ 2(40\cdot 4)=320\\ 4^2=16\\ \hline 91936 \end{array}\] \[\begin{array}{r} \textrm{Negative Parts.}\\ 2(40 \cdot 300)=24000\\ 2(4 \cdot 300) = 2400\\ \hline -26400 \end{array}\] \[\begin{array}{r} 91936\\ -26400\\ \hline 65536 \end{array}\]the square of 256 as before.

#### 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.