Elements of Algebra

366 ELEMENTS PART II. will furnish new ones, by following the methods which have been pointed out. 123. It must be remarked, however, that if we gave ourselves the trouble of deducing new values from the two, which the known case of $x=1$ has furnished, we should arrive at fractions extremely prolix; and we have reason to be surprised that the case, $x=1$, has not rather led us to the other, $x=2$, which is no less evident. This, indeed, is an imperfection of the present method, which is the only mode of proceeding hitherto known. We may, in the same manner, set out from the case $x=2$, in order to find other values. Let us, for this purpose, make $x=2+y$, and it will be required to make a square of the formula, $25+36y+18y^2+3y^3$. Here, if we suppose its root, according to the first method, to be $5+py$, we shall have $25+36y+18y^2+3y^3 = 25+10py+p^2y^2$; and, consequently, $10p=36$, or $p=\frac{18}{5}$: then expunging the terms which destroy each other, and dividing the others by $y^2$, there results $18+3y=p^2=\frac{324}{25}$; consequently, $y=-\frac{42}{25}$, and $x=\frac{8}{25}$; whence it follows, that $1+3x^3$ is a square, whose root is $5+py = -\frac{131}{125}$ or $+\frac{131}{125}$. In the second method, it would be necessary to suppose the root $=5+py+qy^2$, and we should then have $25+36y+18y^2+3y^3 = \{ 25+10py+10qy^2+2pqy^3 +p^2y^2+q^2y^4 \}$; the second and third terms would disappear by making $10p=36$, or $p=\frac{18}{5}$, and $10q+p^2=18$, or $10q=18-\frac{324}{25}=\frac{126}{25}$, or $q=\frac{63}{125}$; and then the other terms, divided by $y^3$, would give $2pq+q^2y=3$, or $q^2y=3-2pq=-\frac{393}{625}$; that is, $y=-\frac{3275}{1323}$, and $x=-\frac{629}{1323}$. 124. This calculation does not become less tedious and difficult, even in the cases where, setting out differently, we can give a general solution; as, for example, when the formula proposed is $1-x-x^2+x^3$, in which we may make, generally, $x=n^2-1$, by giving any value whatever to $n$: for, let $n=2$; we have then $x=3$, and the formula becomes $1-3-9+27=16$. Let $n=3$, we have then $x=8$, and the formula becomes $1-8-64+512=441$, and so on. But it should be observed, that it is to a very peculiar circumstance we owe a solution so easy, and this circumstance is readily perceived by analysing our formula into factors; for we immediately see, that it is divisible by