Elements of Algebra

CHAP. V. OF ALGEBRA. 337 square. Next, if $x$ be a fraction, let us express it by $\frac{t}{u}$, supposing it already reduced to its lowest terms, and that $t$ and $u$ have no common divisor. In order, therefore, that $\frac{3t^2}{u^2} + 2$ may be a square, we must obtain, after multiplying by $u^2$, $3t^2 + 2u^2$ also a square. Now, this is impossible; for the number $u$ is either divisible by 3, or it is not: if it be, $t$ will not be so, for $t$ and $u$ have no common divisor, since the fraction $\frac{t}{u}$ is in its lowest terms. Therefore, if we make $u = 3f$, as the formula becomes $3t^2 + 18f^2$, it is evident that it can be divided by 3 only once, and not twice, as it must necessarily be if it were a square; in fact, if we divide by 3, we obtain $t^2 + 6f^2$. Now, though one part, $6f^2$, is divisible by 3, yet the other, $t^2$, being divided by 3, leaves 1 for a remainder. Let us now suppose that $u$ is not divisible by 3, and see what results from that supposition. Since the first term is divisible by 3, we have only to learn what remainder the second term, $2u^2$, gives. Now, $u^2$ being divided by 3, leaves the remainder 1, that is to say, it is a number of the class $3n + 1$; so that $2u^2$ is a number of the class $6n + 2$; and dividing it by 3, the remainder is 2; consequently, the formula $3t^2 + 2u^2$, if divided by 3, leaves the remainder 2, and is certainly not a square number. 67. We may, in the same manner, demonstrate, that the formula $3t^2 + 5u^2$, likewise can never become a square, nor any one of the following: $3t^2 + 8u^2$, $3t^2 + 11u^2$, $3t^2 + 14u^2$, &c. in which the numbers 5, 8, 11, 14, &c. divided by 3, leave 2 for a remainder. For, if we suppose that $u$ is divisible by 3, and, consequently, that $t$ is not so, and if we make $u=3n$, we shall always be brought to formulæ divisible by 3, but not divisible by 9: and if $u$ were not divisible by 3, and, consequently, $u^2$ a number of the kind $3n + 1$, we should have the first term, $3t^2$, divisible by 3, while the second terms, $5u^2$, $8u^2$, $11u^2$, &c. would have the forms $15n + 5$, $24n + 8$, $33n + 11$, &c. and, when divided by 3, would constantly leave the remainder 2. 68. It is evident that this remark extends also to the general formula, $3t^2 + (3n + 2) \times u^2$, which can never become a square, even by taking negative numbers for $n$. If, for example, we should make $n = -1$, I say, it is im- z