Steps / Historical AI

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

Elements of Algebra

All 627 leaves · leaf 387

Transcription

350 ELEMENTS PART II. and these two series of numbers may be easily continued to any length. 95. It will be observed, however, that here we cannot continue the upper series for $x$, without having the under one in view; but it is easy to remove this inconvenience, and to give a rule, not only for finding the upper series, without knowing the other, but also for determining the latter without the former. The numbers which may be substituted for $x$ succeed each other in a certain progression, such that each term (as, for example, E), may be determined by the two preceding terms C and D, without having recourse to the terms of the second series R and S. In fact, since $E = nS + mD = n(mR + anC) + m(nR + mC) = 2mnR + an²C + m²C$, and $nR = D - mC$, we therefore find $E = 2mD - m²C + an²C$, or $E = 2mD - (m² - an²)C$; or lastly, $E = 2mD - C$, because $m² = an² + 1$, and $m² - an² = 1$; from which it is evident, how each term is determined by the two which precede it. It is the same with respect to the second series; for, since $T = mS + anD$, and $D = nR + mC$, we have $T = mS + an²R + amnC$. Farther, $S = mR + anC$, so that $anC = S - mR$; and if we substitute this value of $anC$, we have $T = 2mS - R$, which proves that the second progression follows the same law, or the same rule, as the first. Let it be required, as an example, to find all the integer numbers, $x$, such, that $2x² - 1 = y²$. We shall first have $f=1$, and $g=1$. Then $m² = 2n² + 1$, if $n=2$, and $m=3$; therefore, since $A = ng + mf = 5$, the first two terms will be 1 and 5; and all the succeeding ones will be found by the formula $E = 6D - C$: that is to say, each term taken six times and diminished by the preceding term, gives the next. So that the numbers $x$ which we require, will form the following series: 1, 5, 29, 169, 985, 5741, &c. This progression we may continue to any length; and if we choose to admit fractional terms also, we might find an infinite number of them by the method which has been already explained*. * See the Appendix to this chapter in the additions by De la Grange, p. 550, et seq.
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