Elements of Algebra

404 ELEMENTS PART II. being laid down, we have to consider what must be the nature of the values of $x$ and $y$, in order that the formula $x^2 + cy^2$ may become a biquadrate. Now, it is composed of the two factors $(x+y\sqrt{-c}) \times (x-y\sqrt{-c})$; and each of these factors must also be a biquadrate of the same kind; therefore we must make $x+y\sqrt{-c}=(p+q\sqrt{-c})^4$, and $x-y\sqrt{-c}=(p-q\sqrt{-c})^4$, whence it follows, that the formula proposed becomes equal to the biquadrate $(p^2+cq^2)^4$. With regard to the values of $x$ and $y$, they are easily determined by the following analysis: $x+y\sqrt{-c}=p^4+4p^3q\sqrt{-c}-6cp^2q^2+c^2q^4-4cpq^3\sqrt{-c}$, $x-y\sqrt{-c}=p^4-4p^3q\sqrt{-c}-6cp^2q^2+c^2q^4+4cpq^3\sqrt{-c}$, whence, $x=p^4-6cp^2q^2+c^2q^4$; and $y=4p^3q-4cpq^3$. 199. So that when $x^2+y^2$ is a biquadrate, because $c=1$, we have $x = p^4-6p^2q^2+q^4$; and $y = 4p^3q-4pq^3$; so that $x^2+y^2=(p^2+q^2)^4$. Suppose, for example, $p=2$, and $q=1$; we shall then find $x=7$, and $y=24$; whence $x^2+y^2=625=5^4$. If $p=3$, and $q=2$, we obtain $x=119$, and $y=120$, which gives $x^2+y^2=13^4$. 200. Whatever be the even power into which it is required to transform the formula $ax^2+cy^2$, it is absolutely necessary that this formula be always reducible to a square; and for this purpose, it is sufficient that we already know one case in which it happens; for we may then transform the formula, as has been seen, into a quantity of the form $t^2+acu^2$, in which the first term $t^2$ is multiplied only by $1$; so that we may consider it as contained in the expression $x^2+cy^2$; and in a similar manner, we may always give to this last expression the form of a sixth power, or of any higher even power. 201. This condition is not requisite for the odd powers; and whatever numbers $a$ and $c$ be, we may always transform the formula $ax^2+cy^2$ into any odd power. Let the fifth, for instance, be demanded; we have only to make $x\sqrt{a}+y\sqrt{-c}=(p\sqrt{a}+q\sqrt{-c})^5$, and $x\sqrt{a}-y\sqrt{-c}=(p\sqrt{a}-q\sqrt{-c})^5$, and we shall evidently obtain $ax^2+cy^2=(ap^2+cq^2)^5$. Farther, as the fifth power of $p\sqrt{a}+q\sqrt{-c}$ is $=a^2p^5\sqrt{a}+5a^2p^4q\sqrt{-c}-10acp^3q^2\sqrt{a}-10acp^2q^3\sqrt{-c}+5c^2pq^4\sqrt{a}+c^2q^5\sqrt{-c}$, we shall, with the same facility, find $x = a^2p^5-10acp^3q^2+5c^2pq^4$, and $y = 5a^2p^4q-10acp^2q^3+c^2q^5$. If it is required, therefore, that the sum of two squares,