Steps / Historical AI

Leonhard Euler · 1770 (1828 English ed.) · Internet Archive

Elements of Algebra

All 627 leaves · leaf 379

Transcription

342 ELEMENTS PART II. But there is frequently an infinite number of cases, in which $x$ may be assigned even in integer numbers; and the determination of those cases shall form the subject of the following chapter. *** CHAP. VI. *Of the Cases in Integer Numbers, in which the Formula $ax^2 + b$ becomes a Square.* 79. We have already shewn [Art. 63], how such formulæ as $a + bx + cx^2$, are to be transformed, in order that the second term may be destroyed; we shall therefore confine our present inquiries to the formula $ax^2 + b$, in which it is required to find for $x$ only integer numbers, which may transform that formula into a square. Now, first of all, such a formula must be possible; for, if it be not, we shall not even obtain fractional values of $x$, far less integer ones. 80. Let us suppose then $ax^2 + b = y^2$; $a$ and $b$ being integer numbers, as well as $x$ and $y$. Now, here it is absolutely necessary for us to know, or to have already found a case in integer numbers; otherwise it would be lost labour to seek for other similar cases, as the formula might happen to be impossible. We shall, therefore, suppose that this formula becomes a square, by making $x = f$, and we shall represent that square by $g^2$, so that $af^2 + b = g^2$, where $f$ and $g$ are known numbers. Then we have only to deduce from this case other similar cases; and this inquiry is so much the more important, as it is subject to considerable difficulties; which, however, we shall be able to surmount by particular artifices. 81. Since we have already found $af^2 + b = g^2$, and likewise, by hypothesis, $ax^2 + b = y^2$, let us subtract the first equation from the second, and we shall obtain a new one, $ax^2 - af^2 = y^2 - g^2$, which may be represented by factors in the following manner; $a(x+f)(x-f) = (y+g)(y-g)$, and which, by multiplying both sides by $pq$, becomes $apq(x+f)(x-f) = pq(y+g)(y-g)$. If we now decompound this equation, by making $ap(x+f) = q(y+g)$, and $q(x-f) = p(y-g)$, we may derive from these two equations values of the two letters $x$ and $y$. The
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