Elements of Algebra

536 ADDITIONS. CHAP. IV. the condition in question, we shall have, in general, $n = n' + \mu c$, $\mu$ being any number whatever. Whence I conclude, that if we substitute successively, in the formula, $a + bx + dx^2 + fx^3 +$, &c. instead of $x$, all the integers positive, or negative, that do not exceed $c/2$, and if we denote by $n'$, $n''$, $n'''$, &c. such of those numbers as will render the quantity, $a + bx + dx^2 +$, &c. divisible by $c$, all the other numbers that do the same, will necessarily be included in the formulæ $n' \pm \mu'c$, $n'' \pm \mu''c$, $n''' \pm \mu'''c$, &c. $\mu'$, $\mu''$, $\mu'''$, &c. being any integer numbers. Various remarks might here be made to facilitate the finding of the numbers $n'$, $n''$, $n'''$, &c. but it is the more unnecessary to enlarge upon this subject, as I have already had occasion to treat of it, in a Memoir published among those of the Academy of Berlin for the year 1768, and entitled *Nouvelle Methode pour resoudre les Problemes indeterminés*. 48. I shall, however, say a word on the method of determining two numbers, $x$ and $y$, so that the fraction $ \frac{ay^m + by^{m-1}x + dy^{m-2}x^2 + fy^{m-3}x^3 +, \&c.}{c} $ may become an integer number, as this investigation will be very useful to us in the sequel. Supposing that $y$ and $x$ must be prime to each other, and farther, that $y$ must be prime to $c$, we may always make $x = ny - cz$; $n$ and $z$ being indeterminate numbers; for, considering $x$, $y$, and $c$, as given numbers, we shall have an equation always resolvible in whole numbers by the method of Chap. III., because $y$ and $c$ have no common measure, by the hypothesis. Now, if we substitute this expression of $x$ in the quantity, $ay^m + by^{m-1}x + dy^{m-2}x^2 +$, &c. it will become, $(a + bn + dn^2 + fn^3 +, \&c.)y^m$ $- (b + 2dn + 3fn^2 +, \&c.)cy^{m-1}z$ $+ (d + 3fn +, \&c.)c^2y^{m-2}z^2$ $ -, \&c. $ and it is evident, that this quantity could not be divisible by $c$, unless the first term, $(a + bn + dn^2 + fn^3 +, \&c.)y^m$ were so, since all the other terms are multiplied by $c$. Therefore, as $c$ and $y$ are supposed to be prime to each other, the quantity $a + bn + dn^2 + fn^3 +$, &c. must itself be divisible by $c$; so that we shall only have to seek, by the method of the preceding Article, all the values of $n$ that can satisfy this