Elements of Algebra

CHAP. VII. ADDITIONS. 577 But the series in question contains only the numbers 1, 15, 2, always repeated; therefore, we may immediately conclude, that the last equation is not resolvible; and, consequently, the equation proposed is not, at least when we take $n = 33$. It only remains, therefore, to try the other value of $n = -33$, which will give $m = \frac{-33}{10} = -3$, $n' = -33 + 3 \times 10 = -3$, $D'' = \frac{9-97}{10} = -7$, $y = -3y' + y''$; so that we shall have the equation transformed into $-7y' + 6y'y'' + 10y''^2 = 1$, which may be reduced to the form, $(7y' - 3y'')^2 - 79y''^2 = -7$, which is similar to the preceding. Whence I conclude, that the given equation absolutely admits of no solution in whole numbers. 84. Scholium. M. Euler, in an excellent Memoir printed in Vol. IX. of the New Commentaries of Petersburg, finds by induction this rule for determining the resolvibility of every equation of the form $x^2 - Ay^2 = B$, when B is a prime number: it is, that the equation must be possible, whenever B shall have the form $4An + r^2$, or $4An + r^2 - A$; but the foregoing example shews this rule to be defective; for 101 is a prime number, of the form $4An + r^2 - A$, making $A = 79$, $n = -4$, and $r = 38$; yet the equation, $x^2 - 79y^2 = 101$, admits of no solution in whole numbers. If the foregoing rule were true, it would follow, that, if the equation $x^2 - Ay^2 = B$ were possible, when B has any value whatever, $b$, it would be so likewise, when we have taken $B = 4An + b$, provided B were a prime number. We might limit this last rule, by requiring $b$ to be also a prime number; but even with this limitation the preceding example would shew it to be false; for we have $101 = 4An + b$, by taking $A = 79$, $n = -2$, and $b = 733$; now, 733 is a prime number, of the form $x^2 - 79y^2$, making $x = 38$, and $y = 3$; yet 101 is not of the same form, $x^2 - 79y^2$. P P