Elements of Algebra

390 ELEMENTS PART II. are both even, because we have already said, that these numbers must not have a common divisor. 169. The two numbers $x$ and $y$ must therefore be prime to each other, and yet the formula $x^2+y^2$ must contain two or more factors. The preceding method does not apply here, because the formula is not resolvible into two rational factors; but the irrational factors, which compose the formula, and which may be represented by the product $(x+y\sqrt{-1}) \times (x-y\sqrt{-1})$, will answer the same purpose. In fact, we are certain, if the formula $x^2+y^2$ have real factors, that these irrational factors must be composed of other factors; because, if they had not divisors, their product could not have any. Now, as these factors are not only irrational, but imaginary; and farther, as the numbers $x$ and $y$ have no common divisor, and therefore cannot contain rational factors; the factors of these quantities must also be irrational, and even imaginary. 170. If, therefore, we wish the formula $x^2+y^2$ to have two rational factors, we must resolve each of the two irrational factors into two other factors; for which reason, let us first suppose $x+y\sqrt{-1} = (p+q\sqrt{-1}) \times (r+s\sqrt{-1})$; and since $\sqrt{-1}$ may be taken *minus*, as well as *plus*, we shall also have $x-y\sqrt{-1}=(p-q\sqrt{-1}) \times (r-s\sqrt{-1})$. Let us now take the product of these two quantities, and we shall find our formula $x^2+y^2 = (p^2+q^2) \times (r^2+s^2)$; that is, it contains the two rational factors $p^2+q^2$, and $r^2+s^2$. It remains, therefore, to determine the values of $x$ and $y$ which must likewise be rational. Now, the supposition we have made, gives $x+y\sqrt{-1}=pr-qs+ps\sqrt{-1}+qr\sqrt{-1}$; and $x-y\sqrt{-1}=pr-qs-ps\sqrt{-1}-qr\sqrt{-1}$. If we add these formulæ together, we shall have $x=pr-qs$; if we subtract them from each other, we find $2y\sqrt{-1}=2ps\sqrt{-1}+2qr\sqrt{-1}$, or $y=ps+qr$. Hence it follows, if we make $x=pr-qs$, and $y=ps+qr$, that our formula $x^2+y^2$ must have two factors, since we find $x^2+y^2=(p^2+q^2) \times (r^2+s^2)$. If, after this, a greater number of factors be required, we have only to assign, in the same manner, such values to $p$ and $q$, that $p^2+q^2$ may have two factors; we shall then have three