Elements of Algebra

336 ELEMENTS PART II. $cz^2/z^2$, that is to say, $c$, be one. So that when $c$ is not a square, the formula $cz^2$ can by no means become a square; and on the contrary, if $c$ be itself a square, $cz^2$ will also be a square, whatever number be assumed for $z$. 65. If we wish to consider other cases, we must have recourse to what has been already said on the subject of different kinds of numbers, considered with relation to their division by other numbers. We have seen, for example, that the divisor $3$ produces three different kinds of numbers. The first comprehends the numbers which are divisible by $3$, and may be expressed by the formula $3n$. The second kind comprehends the numbers which, being divided by $3$, leave the remainder $1$, and are contained in the formula $3n + 1$. To the third class belong numbers which, being divided by $3$, leave $2$ for the remainder, and which may be represented by the general expression $3n + 2$. Now, since all numbers are comprehended in these three formulæ, let us therefore consider their squares. First, if the question relate to a number included in the formula $3n$, we see that the square of this quantity being $9n^2$, it is divisible not only by $3$, but also by $9$. If the given number be included in the formula $3n + 1$, we have the square $9n^2 + 6n + 1$, which, divided by $3$, gives $3n^2 + 2n$, with the remainder $1$; and which, consequently, belongs to the second class, $3n + 1$. Lastly, if the number in question be included in the formula $3n + 2$, we have to consider the square $9n^2 + 12n + 4$; and if we divide it by $3$, we obtain $3n^2 + 4n + 1$, and the remainder $1$; so that this square belongs, as well as the former, to the class $3n + 1$. Hence it is obvious, that square numbers are only of two kinds with relation to the number $3$; for they are either divisible by $3$, and in this case are necessarily divisible also by $9$; or they are not divisible by $3$, in which case the remainder is always $1$, and never $2$; for which reason, no number contained in the formula $3n + 2$ can be a square. 66. It is easy, from what has just been said, to shew, that the formula $3x^2 + 2$ can never become a square, whatever integer, or fractional number, we choose to substitute for $x$. For, if $x$ be an integer number, and we divide the formula $3x^2 + 2$ by $3$, there remains $2$; therefore it cannot be a