Elements of Algebra

546 ADDITIONS. CHAP. V. each other, whereas $q'$ and $p$ cannot be), and we shall have to make a square of the quantity $-5r^2 + 3s^2$; so that calling the root $z'$, we shall have $-5r^2 + 3s^2 = z'^2$, and thence $-5r^2 = z'^2 - 3s^2$. We shall, therefore, take $z' = ms + 5s$, and $m$ must be an integer number not $7\frac{1}{2}$, and such, that $m^2 - 3$ may be divisible by 5. Now, this is impossible; for we can only take $m=1$, or $m=2$, which gives $m^2 - 3 = -2$, or $=1$. From this, therefore, we may conclude that the problem is not resolvible; that is to say, it is impossible for the formula $23y^2 - 5$ ever to become a square, whatever number we substitute for $y$*. 59. *Corollary*. If we had a quadratic equation, with two unknown quantities, such as $a + bx + cy + dx^2 + exy + fy^2 = 0$, and it were proposed to find rational values of $x$ and $y$ that would satisfy the conditions of this equation, we might do this, when it is possible, by the method already explained. Taking the value of $y$ in $x$, we have $2fy + ex + c = \sqrt{((c-ex)^2 - 4f(a+bx+dx^2))}$; or, making $\alpha = c^2 - 4af$, $\beta = 2ce - 4bf$, $\gamma = e^2 - 4df$, $2fy + ex + c = \sqrt{(\alpha + \beta x + \gamma x^2)}$; the question will be reduced to finding the values of $x$, that will render rational the radical quantity $\sqrt{(\alpha + \beta x + \gamma x^2)}$. 60. *Scholium*. I have already considered this subject, rather differently, in the Memoirs of the Academy of Sciences at Berlin, for the year 1767, and, I believe, first gave a direct method, without the necessity of trial, for solving indeterminate problems of the second degree. The reader, who wishes to investigate this subject fully, may consult those Memoirs; where he will, in particular, find new and important remarks on the investigation of such integer numbers as, when taken for $n$, will render $n^2 - B$ divisible by $A$, $A$ and $B$ being given numbers. * The impossibility of the formula $23y^2 - 5 = z^2$ is readily demonstrated: for $y^2$ must be of one of the forms $4n$, or $4n+1$. In the first case, $23y^2 - 5$ is of the form $23 \times 4n - 5$, which is the same as $4n-1$, and this is an impossible form for square numbers. In the second case, $23y^2 - 5$ is of the form $23 \times (4n+1) - 5$, which is the same as $4n-18$, or $4n-2$, and this again is an impossible form for square numbers. Therefore, the formula $23y^2 - 5 = z^2$ is always impossible. B.