Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 607
Transcription
570
ADDITIONS.
CHAP. VII.
$\Delta'$, we should in like manner seek the corresponding values of $M$ and $N$, which we shall here express by $M'$ and $N'$, and all the values of the exponent $m$, that will satisfy the condition proposed, will be contained in the formula $N' + \lambda'M'$; $\lambda'$ being any whole number whatever. So that we shall only have to seek the values, which we must give to the whole numbers $\lambda$ and $\lambda'$, in order that we may have $N + \lambda M = N' + \lambda' M'$, or $M\lambda - M'\lambda' = N' - N$, an equation resolvible by the method of Art. 42.
It is easy to apply what we have just now said to the case of Art. 77, where the given expressions have the form, $\alpha t + \beta u + \gamma$, $\alpha' t + \beta' u + \gamma'$, and the divisors are $\delta$ and $\delta'$.
We must only recollect to take the numbers $t$ and $u$, successively, positive and negative, in order to have all the cases that are possible.
80. Scholium. If the equation proposed for resolution, in whole numbers, were of the form $ar^2 + 2brs + cs^2 = f$, we might immediately apply to it the method of Art. 65; for, 1st, it is evident that $r$ and $s$ could have no common divisor, unless the number $f$ were at the same time divisible by the square of that divisor; so that we may always reduce the question to the case, in which $r$ and $s$ shall be prime to each other. 2dly, It is evident, also, that $s$ and $f$ could have no common divisor, unless that divisor were one also of the number $a$, supposing $r$ prime to $s$; so that we may also reduce the question to the case, in which $s$ and $f$ shall be prime to each other. (See Art. 64).
Now, $s$ being supposed prime to $f$, and to $r$, we may make $r = ns - fz$; and, in order that the equation may be resolvible in whole numbers, there must be a value of $n$, positive or negative, not greater than $\frac{f}{2}$, which may render the quantity $an^2 + 2bn + c$ divisible by $f$. This value being substituted for $n$, the whole equation will become divisible by $f$, and will be found reduced to the case of Art. 66, et seq.
It is easy to perceive, that the same method may serve for reducing every equation of the form, $ar^m + br^{m-1}s + cr^{m-2}s^2 +$, &c. $+ ks^m = f$, $a, b, c,$ &c. being given whole numbers, and $r$ and $s$ being two indeterminate numbers, which must likewise be integers, in another similar equation, but in which the whole known term is unity, and then we may apply to it the general method of Chap. 2. See the Scholium of Art. 30.
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