Elements of Algebra

338 ELEMENTS PART II. possible for the formula $3t^2 - u^2$ to become a square. This is evident, if $u$ be divisible by 3: and if it be not, then $u^2$ is a number of the kind $3n+1$, and our formula becomes $3t^2 - 3n - 1$, which, being divided by 3, gives the remainder $-1$, or $+2$; and in general, if $n = -m$, we obtain the formula $3t^2 - (3m-2)u^2$, which can never become a square. 69. So far, therefore, are we led by considering the divisor 3; if we now consider 4 also as a divisor, we see that every number may be comprised in one of the four following formulæ: $4n, 4n+1, 4n+2, 4n+3$. The square of the first of these classes of numbers is $16n^2$; and, consequently, it is divisible by 16. That of the second class, $4n+1$, is $16n^2+8n+1$; which, if divided by 8, the remainder is 1; so that it belongs to the formula $8n+1$. The square of the third class, $4n+2$, is $16n^2+16n+4$; which, if we divide by 16, there remains 4; therefore this square is included in the formula $16n+4$. Lastly, the square of the fourth class, $4n+3$, being $16n^2+24n+9$, it is evident that dividing by 8 there remains 1. 70. This teaches us, in the first place, that all the even square numbers are either of the form $16n$, or $16n+4$; and, consequently, that all the other even formulæ, namely, $16n+2, 16n+6, 16n+8, 16n+10, 16n+12, 16n+14$, can never become square numbers. Secondly, that all the odd squares are contained in the formula $8n+1$; that is to say, if we divide them by 8, they leave a remainder of 1. And hence it follows, that all the other odd numbers, which have the form either of $8n+3$, or of $8n+5$, or of $8n+7$, can never be squares. 71. These principles furnish a new proof, that the formula $3t^2+2u^2$ cannot be a square. For, either the two numbers $t$ and $u$ are both odd, or the one is even and the other odd. They cannot be both even, because in that case they would, at least, have the common divisor 2. In the first case, therefore, in which both $t^2$ and $u^2$ are contained in the formula $8n+1$, the first term $3t^2$, being divided by 8, would leave the remainder 3, and the other term $2u^2$ would leave the remainder 2; so that the whole remainder would be 5: consequently, the formula in question cannot be a square. But, if the second case be supposed, and $t$ be even, and $u$ odd, the first term $3t^2$ will be divisible by 4, and the