Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 572
Transcription
CHAP. IV.
ADDITIONS.
535
Now, since $y = - \frac{p}{q}$, we shall also have $p = -qy$; so that by substituting this value of $p$, we shall get
$A - Byq + Cq + Dy^2q - Eyq^2 + Fq^2 +, \text{\&c.} = 0$,
where all the terms are multiplied by $q$, except the first, $A$; therefore the number $A$ must be divisible by the number $q$, otherwise it would be impossible for the numbers $q$ and $y$ to be both integers.
We shall therefore seek all the divisors of the known integer number $A$, and shall successively take each of these divisors for $q$; from each of which suppositions we shall have a determinate equation in $x$, the integer and rational roots of which, if it have any, will be found by the known methods; then substituting these roots for $x$, we shall see whether the values of $p$ and $q$, which result, are such, that $\frac{p}{q}$ may be an integer number. By these means, we shall certainly find all the integer values of $x$, which may likewise give integer values of $y$ in the equation proposed.
Hence we see, that the number of integer solutions of such equations must always be limited; but there is one case which must be excepted, and which does not fall under the preceding method.
47. This case is when there are no coefficients $e, g, k$, &c. So that we have simply,
$y = \frac{a+bx+dx^2+fx^3+hx^4+, \text{\&c.}}{c}$
In order to find all the values of $x$, that will render the quantity $a + bx + dx^2 + fx^3 + hx^4 +, \text{\&c.}$ divisible by the quantity $c$, we must proceed as follows. Suppose we have already found an integer, $n$, which satisfies this condition; it is evident that every number of the form $n \pm \mu c$ will likewise satisfy it, $\mu$ being any integer number; farther, if $n$ is $> \frac{c}{2}$ (abstracting from the signs of $n$ and $c$), we may always determine the number $\mu$, and the sign which precedes it, so that the number $n \pm \mu c$ may become $\le \frac{c}{2}$; and it is easy to perceive that this could only be done in one way, the values of $n$ and $c$ being given; wherefore, if we express by $n'$ that value of $n \pm \mu c$, which is $\le \frac{c}{2}$, and which satisfies
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