Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 574
Transcription
CHAP. V.
ADDITIONS.
537
condition, and then we shall have, in general, $x = ny - az$, $z$ being any integer number whatever.
It is proper to observe, that although we have supposed the numbers $x$ and $y$ to be prime to each other, as well as the numbers $y$ and $c$, our solution is still no less general; for if $x$ and $y$ had a common measure $\alpha$, we should only have to substitute $ax'$ and $ay'$, instead of $x$ and $y$, and should then consider $x'$ and $y'$ as prime to each other; likewise, if $y'$ and $c$ were a common measure $\beta$, we might put $\beta y''$, instead of $y'$, and consider $y''$ and $c$ as prime to each other.
CHAP. V.
*A direct and general method for finding the values of $x$, that will render Quantities of the form $\sqrt{(a+bx+cx^2)}$, Rational, and for resolving, in Rational Numbers, the indeterminate Equations of the second Degree, which have two unknown Quantities, when they admit of Solutions of this kind.*
[APPENDIX TO CHAP. IV.]
49. I suppose first that the known numbers $a, b, c$, are integers; for if they were fractions, we should only have to reduce them to a common square denominator, and then it is evident, that we might always abstract from their denominator; but with respect to the number $x$, we shall suppose that it may be integer, or fractional, and shall see, in what follows, how the question is to be resolved, when we admit only integer numbers.
Let then $\sqrt{(a + bx + cx^2)} = y$, and we shall have $2cx + b = \sqrt{(4cy^2 + b^2 - 4ac)}$; so that the difficulty will be reduced to rendering rational the quantity,
$\sqrt{(4cy^2 + b^2 - 4ac)}$.
50. Let us suppose, therefore, in general, that we have to make rational the quantity $\sqrt{(AY^2 + B)}$; that is to say, to make $AY^2 + B$ equal to a square, $A$ and $B$ being given integer numbers positive or negative, and $y$ an indeterminate number, which must be rational.
It is evident that if one of the numbers $A$, or $B$, were $1$, or any other square, the problem would be resolvible by the known methods of Diophantus, which are detailed in
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