Translation:Demonstratio nova altera theorematis omnem functionem algebraicam rationalem integram unius variabilis in factores reales primi vel secundi gradus resolvi posse

1.
Although the proof of the theorem on the resolution of algebraic integral functions into factors, which I presented in a paper sixteen years ago, seems to leave nothing to be desired in terms of both rigor and simplicity, I hope that geometers will not be ungrateful if I return to the same most serious question again, and attempt to construct another demonstration from entirely different principles, which will be no less rigorous. Indeed, that previous proof relies at least in part on geometric considerations: on the contrary, the one which I undertake to explain here will be based solely on analytic principles. I have reviewed the most notable analytic methods through which other geometers had attempted to prove our theorem up to that time, and I have extensively exposed the defects they suffered from. The most serious and truly fundamental defect common to all those efforts, as well as to more recent ones that have come to my attention, I declared however not to be inevitable in an analytic demonstration. It is now up to the experts to judge whether the trust once given has been fully justified through these new efforts.

2.
Certain preliminary considerations will precede the main discussion, both to ensure that nothing appears to be lacking, and because the treatment itself may shed new light on even those matters which have already been considered by others. First of all, we will deal with the highest common divisor of two integral algebraic functions of one indeterminate. It should be noted here that we are only concerned with integral functions: from two such functions, if their product is formed, each of them is called a divisor of this product. The order of the divisor is determined by the exponent of the indeterminate in the highest power it contains, without any consideration of numerical coefficients. That which applies to the common divisors of functions can be concluded more briefly, since it is entirely analogous to that which applies to the common divisors of numbers.

Let two functions $Y,$ $Y^{\prime}$  of the indeterminate $x$  be given, where the former is of higher order or at least not of lower order than the latter. We then form the following equations,

$$\begin{array}{c} Y = q Y^{\prime} + Y^{\prime \prime} \\ Y^{\prime} = q^{\prime} Y^{\prime \prime} + Y^{\prime \prime \prime} \\ Y^{\prime \prime} = q^{\prime \prime} Y^{\prime \prime \prime} + Y^{\prime \prime \prime \prime} \\ \text{etc.} \text{ up to } \\ Y^{(\mu-1)} = q^{(\mu-1)} Y^{(\mu)} \end{array}$$

namely by the rule that first $Y$ is divided in the usual way by $Y^{\prime};$  then $Y^{\prime}$  by the remainder of the first division $Y^{\prime \prime},$  which will be of lower order than $Y^{\prime};$  then again the first remainder by the second $Y^{\prime \prime \prime},$  and so on, until we arrive at a division without remainder, which must necessarily happen, as the order of the functions $Y^{\prime},$  $Y^{\prime \prime},$  $Y^{\prime \prime \prime},$  etc., continually decreases. It is scarcely necessary to remark that these functions, as well as the quotients $q,$ $q^{\prime},$  $q^{\prime \prime},$  etc., are integral functions of $x.$  With these preliminaries, it is clear that

I. by going back from the last of these equations to the first, that the function $Y^{(\mu)}$ is a divisor of each of the preceding ones, and consequently a certain common divisor of the given functions $Y,$  ${Y}^{\prime}.$

II. by proceeding from the first equation to the last, any common divisor of the functions ${Y},$ ${Y}^{\prime}$  also measures each of the following ones, and therefore also the last $Y^{(\mu)}.$  Therefore, the functions $Y,$  $Y^{\prime}$  cannot have any common divisor of higher order than $Y^{(\mu)},$  and every common divisor of the same order as $Y^{(\mu)},$  such as ${Y}^{(\mu)},$  will be in the same ratio to this as one number to another, wherefore it must itself be considered as the highest common divisor.

III. If $Y^{(\mu)}$ is of order 0, i.e., a number, then no function of the indeterminate $x$  properly so called can measure the functions $Y,$  ${Y}^{\prime}{:}$  in this case, therefore, it must be said that these functions do not have a common divisor.

IV. If we extract the penultimate equation from our equations, and from this we eliminate $Y^{(\mu-1)}$ by means of the equation before the penultimate one; then again we eliminate ${Y}^{(\mu-2)}$  by means of the preceding equation, and so on, then we will have

$$\begin{aligned} Y^{(\mu)} & =+k Y^{(\mu-2)}-k^{\prime} Y^{(\mu-1)} \\ & =-k^{\prime} Y^{(\mu-3)}+k^{\prime \prime} Y^{(\mu-2)} \\ & =+k^{\prime \prime} Y^{(\mu-4)}-k^{\prime \prime \prime} Y^{(\mu-3)} \\ & =-k^{\prime \prime \prime} Y^{(\mu-5)}+k^{\prime \prime \prime \prime} Y^{(\mu-4)} \\ & \text{ etc., } \end{aligned}$$

provided we suppose the functions $k,$ $k^{\prime},$  $k^{\prime \prime},$  etc., to be formed according to the following rule:

$$\begin{aligned} & k=1 \\ & k^{\prime}=q^{(\mu-2)} \\ & k^{\prime \prime}=q^{(\mu-3)} k^{\prime}+k \\ & k^{\prime \prime \prime}=q^{(\mu-4)} k^{\prime \prime}+k^{\prime} \\ & k^{\prime \prime \prime \prime}=q^{(\mu-5)} k^{\prime \prime \prime}+k^{\prime \prime} \\ & \text{ etc. } \end{aligned}$$

Therefore, we will have

$$\pm k^{(\mu-2)} Y \mp k^{(\mu-1)} Y^{\prime}=Y^{(\mu)}$$

with the upper signs holding when $\mu$ is even and the lower signs holding when it is odd. In the case where $Y$ and $Y^{\prime}$  do not have a common divisor, we can find in this way two functions $Z,$  $Z^{\prime}$  of the indeterminate $x$  such that

$$Z Y+Z^{\prime} Y^{\prime}=1.$$

V. This proposition manifestly also holds in reverse, namely, if the equation

$$Z Y+Z^{\prime} Y^{\prime}=1$$

can be satisfied, such that $Z,$ $Z^{\prime}$  are integral functions of the indeterminate $x,$  then the functions $Y$  and $Y^{\prime}$  cannot have a common divisor.

3.
Another preliminary discussion will concern the transformation of symmetric functions. Let $a,$ $b,$  $c$  etc. be indeterminate quantities, their multitude being $m,$  and let us denote by $\lambda^{\prime}$  the sum of these quantities, by $\lambda^{\prime \prime}$  the sum of their products taken two at a time, by $\lambda^{\prime \prime \prime}$  the sum of their products taken three at a time, etc., so that from the expansion of the product

$$(x-a)(x-b)(x-c) \ldots$$

arises

$$x^{m}-\lambda^{\prime} x^{m-1}+\lambda^{\prime \prime} x^{m-2}-\lambda^{\prime \prime \prime} x^{m-3}+\text{etc. }$$

It follows that the quantities $\lambda^{\prime},$ $ \lambda^{\prime \prime},$  $ \lambda^{\prime \prime \prime},$  etc., are symmetric functions of the indeterminates $a,$  $b,$  $c,$  etc., i.e., such that these indeterminates occur in them in the same manner, or more clearly, such that they are not changed by any permutation of the indeterminates. It is clear that, in general, any integral function of the quantities $\lambda^{\prime},$ $ \lambda^{\prime \prime},$  $ \lambda^{\prime \prime \prime},$  etc., (whether it involves only these indeterminates or still others independent of $a,$  $b,$  $c,$  etc.) will be a symmetric integral function of the indeterminates $a,$  $b,$  $c,$  etc.

4.
The inverse theorem is somewhat less obvious. Let $\rho$ be a symmetric function of the indeterminates $a,$  $b,$  $c,$  etc., which will thus be composed of a certain number of terms of the form

$$M a^{\alpha} b^{\beta} c^{\gamma} \ldots$$

where $\alpha,$ $\beta,$  $\gamma,$  etc. are non-negative integers, and $M$  is a coefficient that is either determinate or at least does not depend on $a,$  $b,$  $c,$  etc. (if it happens that indeterminates besides $a,$  $b,$  $c,$  etc., are involved in the function $\rho$ ). Let us first of all establish a certain order among these terms, to which end we will initially arrange the indeterminates $a,$ $b,$  $c,$  etc., in a definite order, entirely arbitrary in itself, e.g. such that $a$  occupies the first place, $b$  the second, $c$  the third, etc. Then, out of two terms

$M a^{\alpha} b^{\beta} c^{\gamma} \ldots$ and $M a^{\alpha^{\prime}} b^{\beta^{\prime}} c^{\gamma^{\prime}} \ldots$ |undefined

we will assign a higher order to the former than to the latter if either

$\alpha>\alpha^{\prime},$ or $\alpha=\alpha^{\prime}$ and $\beta>\beta^{\prime},$  or $\alpha=\alpha^{\prime},$  $\beta=\beta^{\prime}$  and $\gamma>\gamma^{\prime},$  or etc.,

i.e. if among the differences $\alpha-\alpha^{\prime},$ $\beta-\beta^{\prime},$  $\gamma-\gamma^{\prime},$  etc. the first non-vanishing one turns out to be positive. Therefore, since terms of the same order differ only with respect to the coefficient $M,$ and thus can be merged into one term, we assume that each term of the function $\rho$  belongs to different orders.

Next, we observe that if $M a^{\alpha} b^{\beta} c^{\gamma} \ldots$ has the highest order among all the terms of the function $\rho,$  then $\alpha$  must be greater than, or at least not less than, $\beta.$  For if $\beta>\alpha,$  then the term $M a^{\beta} b^{\alpha} c^{\gamma} \ldots,$  which the function $\rho,$  being symmetric, also involves, would be of higher order than $M a^{\alpha} b^{\beta} c^{\gamma} \ldots,$  contrary to the hypothesis. Similarly, $b$ will be greater than, or at least not less than, $\gamma;$  furthermore, $\gamma$  will be not less than the subsequent exponent $\delta,$  etc.; thus, each of the differences $\alpha-\beta,$  $\beta-\gamma,$  $\gamma-\delta,$  etc., will be non-negative integers.

Secondly, let us consider that if a product is formed from any number of indeterminate integral functions of $a,$ $b,$  $c$  etc., then the highest term of this product must necessarily be the product of the highest terms of those factors. It is equally clear that the highest terms of the functions $\lambda^{\prime},$ $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc., respectively, are $a,$  $a b,$  $a b c$  etc. Hence it follows that the highest term produced by the product

$$p=M \lambda^{\prime \alpha-\beta} \lambda^{\prime \prime \beta-\gamma} \lambda^{\prime \prime \prime \gamma-\delta} \ldots$$

will be $M a^{\alpha} b^{\beta} c^{\gamma} \ldots;$ hence by setting $\rho-p=\rho^{\prime},$  the highest term of the function $\rho^{\prime}$  will certainly be of lower order than the highest term of the function $\rho.$  Moreover, it is clear that $p,$  and therefore $\rho^{\prime},$  will become an integral symmetric function of $a,$  $b,$  $c$  etc. Therefore, treating $\rho^{\prime}$  just as $\rho$  was treated before, it will be split into $p^{\prime}+\rho^{\prime \prime},$  so that $p^{\prime}$  is a product of powers of $\lambda^{\prime},$  $ \lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc., with coefficients that are either determinate or at least independent of $a,$  $b,$  $c$  etc., and $\rho^{\prime \prime}$  is an integral symmetric function of $a,$  $b,$  $c$  etc. whose highest term belongs to a lower order than the highest term of the function $p^{\prime}.$  Continuing in the same way, it is clear that $\rho$  will eventually  be reduced to the form $p+p^{\prime}+p^{\prime \prime}+p^{\prime \prime \prime}$  etc., i.e. it will be transformed into an integral function of $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc.

5.
The theorem demonstrated in the preceding article can also be stated as follows: Given any symmetric integral function of the indeterminates $a,$ $b,$  $c$  etc., $\rho,$  another integral function of an equal number of other indeterminates $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., can be assigned, such that through substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc., it passes into $\rho.$  Moreover, it can be easily shown that this can only be done in one way. For suppose that there are two different functions $r$ and $r^{\prime}$  of the indeterminates $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., such that upon substituting $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc. these functions yield the same function of $a,$  $b,$  $c$  etc. Then $r-r^{\prime}$  will be a function of $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which does not vanish by itself, but which is identically destroyed after those substitutions. However, it is easy to see that this is absurd, by considering that $r-r^{\prime}$ must necessarily be composed of a certain number of parts of the form

$$M l^{\prime \alpha} l^{\prime \prime \beta} l^{\prime \prime \prime \gamma} \ldots$$

whose coefficients $M$ do not vanish, and which are different with respect to their exponents, so that the highest order terms coming from each of these parts are of the form

$$M a^{\alpha+\beta+\gamma+\text{etc.}} b^{\beta+\gamma+\text{etc.}} c^{\gamma+\text{etc.}} \ldots$$

and therefore must have different orders, so that the highest order term cannot be destroyed in any way.

Moreover, the calculation of such transformations can be significantly abbreviated by several methods, which we do not dwell on here, since for our purpose only the possibility of the transformation is sufficient.

6.
Consider the product of $m(m-1)$ factors

$$\begin{aligned} & (a-b)(a-c)(a-d) \ldots \\ \times & (b-a)(b-c)(b-d) \ldots \\ \times & (c-a)(c-b)(c-d) \ldots \\ \times & (d-a)(d-b)(d-c) \ldots \\ & \text{ etc. } \end{aligned}$$

which we will denote by $\pi.$ Since it involves the indeterminates $a,$  $b,$  $c$  etc. symmetrically, let us assume that it has been reduced to the form of a function of $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc. Let this function transform into $p,$  if we substitute $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc. in place of $\lambda^{\prime}\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc.,  Having done this, we will call $p$  the determinant of the function

$$y=x^{m}-l^{\prime} x^{m-1}+l^{\prime \prime} x^{m-2}-l^{\prime \prime \prime} x^{m-3}+\text{etc.}$$

So, for $m=2,$ we have

$$p=-l^{\prime 2}+4 l^{\prime \prime}$$

Similarly for $m=3,$ we find

$$p=-l^{\prime 2} l^{\prime \prime 2}+4 l^{\prime 3} l^{\prime \prime \prime}+4 l^{\prime \prime 3}-18 l^{\prime} l^{\prime \prime} l^{\prime \prime \prime}+27 l^{\prime \prime \prime 2}$$

The determinant of the function $y$ is therefore a function of coefficients $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which through the substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc., is transformed into the product of all differences between pairs of the quantities $a,$  $b,$  $c$  etc. In the case where $m=1,$  i.e. where only one indeterminate $a$  is present, and thus there are no differences at all, it will be convenient to adopt the number $1$  as the determinant of the function $y.$

In fixing the notion of determinant, the coefficients of the function $y$ should be regarded as indeterminate quantities. The determinant of a function with determined coefficients

$$Y=x^{m}-L^{\prime} x^{m-1}+L^{\prime \prime} x^{m-2}-L^{\prime \prime \prime} x^{m-3}+\text{etc. }$$

will be a determined number $P,$ namely the value of the function $p$  when $l^{\prime}=L^{\prime},$  $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime}$  etc. Therefore, if we suppose that $Y$  can be resolved into simple factors, as

$$Y=(x-A)(x-B)(x-C) \ldots$$

or that $Y$ arises from

$$\upsilon=(x-a)(x-b)(x-c)$$

by setting $a=A,$ $b=B,$  $c=C$  etc., and thus by the same substitutions by which $\lambda^{\prime}, \lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime}$  etc., become $L^{\prime},$  $L^{\prime \prime},$  $L^{\prime \prime}$  etc., it is clear that $P$  will be equal to the product of factors

$$\begin{aligned} & (A-B)(A-C)(A-D) \ldots \\ \times & (B-A)(B-C)(B-D) \cdots \\ \times & (C-A)(C-B)(C-D) \cdots \\ \times & (D-A)(D-B)(D-C) \cdots \\ &\text{etc.} \end{aligned}$$

Therefore, it is clear that if $P=0,$ at least two of the quantities $A,$  $B,$  $C$  etc., must be equal; on the contrary, if $P \neq 0,$  then $A,$  $B,$  $C$  etc. must all be different. Now we observe that if we set $\frac{\operatorname{d} Y}{\operatorname{d} x}=Y^{\prime},$ or

$$Y^{\prime}=m x^{m-1}-(m-1) L^{\prime} x^{m-2}+(m-2) L^{\prime \prime} x^{m-3}-(m-3) L^{\prime \prime \prime} x^{m-4}+\text{etc. }$$

then we have

$$\begin{aligned} Y^{\prime}=\phantom{+}& (x-B)(x-C)(x-D) \ldots \\ +&(x-A)(x-C)(x-D) \ldots \\ +&(x-A)(x-B)(x-D) \ldots \\ +&(x-A)(x-B)(x-C) \ldots \\ +& \text{ etc. } \end{aligned}$$

Therefore, if two of the quantities $A,$ $B,$  $C$  etc., are equal, e.g. $A=B,$  then $Y^{\prime}$  will be divisible by $x-A,$  meaning that $Y$  and $Y^{\prime}$  will have a common divisor $x-A.$  Vice versa, if it is supposed that $Y^{\prime}$  and $Y$  have a common divisor, then necessarily $Y^{\prime}$  must involve some simple factor from among $x-A,$  $x-B,$  $x-C$  etc., e.g. the first $x-A,$  which clearly cannot happen unless $A$  is equal to one of the other $B,$  $C,$  $D$  etc. From all this, we obtain two theorems:

I. If the determinant of the function $Y$ becomes $=0,$  then certainly $Y$  and $Y^{\prime}$  have a common divisor, and therefore, if $Y$  and $Y^{\prime}$  do not have a common divisor, the determinant of the function $Y$  cannot be $=0.$ 

II. If the determinant of the function $Y$ is not $=0,$  then certainly $Y$  and $Y^{\prime}$  cannot have a common divisor; or, if $Y$  and $Y^{\prime}$  have a common divisor, the determinant of the function $Y$  must necessarily be $=0.$ 

7.
It should be noted that the entire force of this very simple demonstration relies on the assumption that the function $Y$ can be resolved into simple factors: which assumption, in this place, where the general demonstration of this resolvability is discussed, would be nothing but begging the question. And yet not all have avoided fallacies entirely similar to this, who have attempted analytic demonstrations of the principal theorem, the specious illusion of whose origin we have already observed, as all have sought only the form of the roots of equations, whereas it should have been necessary to demonstrate their existence. However, enough has been said already, in the previously cited commentary, about such a method of proceeding, which deviates too much from rigor and clarity. Therefore, we will now place the theorems in the preceding article, of which we cannot afford to be without at least one for our purpose, on a more solid foundation. We will start from the second, this being the easier one.

8.
Let us denote by $\rho$ the function

$$\begin{aligned} & \frac{\pi(x-b)(x-c)(x-d) \dots \dots}{(a-b)^{2}(a-c)^{2}(a-d)^{2} \dots \dots} \\ + & \frac{\pi(x-a)(x-c)(x-d) \dots \dots}{(b-a)^{2}(b-c)^{2}(b-d)^{2} \dots \dots} \\ + & \frac{\pi(x-a)(x-b)(x-d) \dots \dots}{(c-a)^{2}(c-b)^{2}(c-d)^{2} \dots \dots} \\ + & \frac{\pi(x-a)(x-b)(x-c) \dots \dots}{(d-a)^{2}(d-b)^{2}(d-c)^{2} \dots \dots} \\ + & \text{ etc. } \end{aligned}$$

which, since $\pi$ is divisible by each of the denominators, becomes an integral function of the variables $x,$  $a,$  $b,$  $c$  etc. Let us further set $\frac{\operatorname{d} \upsilon}{\operatorname{d} x}=\upsilon^{\prime},$  so that we have

$$\begin{aligned} \upsilon^{\prime}=\phantom{+} &(x-b)(x-c)(x-d) \dots \dots \\ +&(x-a)(x-c)(x-d) \dots \dots\\ +&(x-a)(x-b)(x-d) \dots \dots\\ +&(x-a)(x-b)(x-c) \dots \dots\\ +&\text{ etc. } \end{aligned}$$

Clearly, for $x=a,$ we have $\rho \upsilon^{\prime}=\pi,$  hence we conclude that the function $\pi$  - $\rho \upsilon^{\prime}$  is indefinitely divisible by $x-a,$  and similarly by $x-b,,$  $x-c,$  etc., as well as by the product $v.$  Therefore, if we set

$$\frac{\pi-\rho\upsilon^{\prime}}{\upsilon}=\sigma$$

then $\sigma$ will be an integral function of the variables $x,$  $a,$  $b,$  $c$  etc., and indeed, like $\rho,$  it will symmetric with respect to the variables $a,$  $b,$  $c,$  etc. Consequently, two integral functions $r,$  $s,$  of the variables $x,$  $l^{\prime}, l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., can be derived, which upon substituting $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime},$  etc. are transformed into $\rho,$  $\sigma,$  respectively. Therefore, following the analogy, if the function

$$m x^{m-1}-(m-1) l^{\prime} x^{m-2}+(m-2) l^{\prime \prime} x^{m-3}-(m-3) l^{\prime \prime \prime} x^{m-4}+\text{etc.}$$

i.e. the differential quotient $\frac{\operatorname{d} y}{\operatorname{d} x},$ is denoted by $y^{\prime},$  then since $y^{\prime}$  is transformed into $v^{\prime}$  by the same substitutions, it is evident that $p-s y-r y^{\prime}$  is transformed into $\pi- \sigma \upsilon-\rho \upsilon^{\prime},$  i.e. into $0,$  and therefore must vanish identically (art. 5): thus, we have the identity

$$p=s y+r y^{\prime}$$

and hence, if we suppose that substituting $l^{\prime}=L^{\prime},$ $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime}$  etc. produces $r=R,$  $s=S,$  then we have the identity

$$P=S Y+R Y^{\prime}$$

where since $S,$ $R$  are integral functions of $x,$  and $P$  is a determined quantity or number, it is evident that $Y$  and $Y^{\prime}$  cannot have a common divisor unless $P=0.$  This is precisely the second theorem of art. 6.

9.
We conclude the demonstration of the previous theorem by showing that, in the case where ${Y}$ and ${Y}^{\prime}$  do not have a common divisor, it cannot be that $P=0.$  To this end, we first, following the instructions of article 2, find two integral functions $f x$  and $\varphi x$  of the variable $x,$  such that we have the identity

$$f x. Y+\varphi x. Y^{\prime}=1$$

which we write as

$$f x. \upsilon+\varphi x. \upsilon^{\prime}=1+f x. (\upsilon-Y)+\varphi x. \frac{\operatorname{d}(\upsilon-Y)}{\operatorname{d} x}$$

or equivalently, since we have

$$\begin{aligned} \upsilon^{\prime} =&(x-b)(x-c)(x-d) \ldots \\ &+(x-a). \frac{\operatorname{d}[(x-b)(x-c)(x-d) \ldots]}{\operatorname{d} x} \end{aligned}$$

in the form

$$\begin{aligned} &\phantom{+.} \varphi x. (x-b)(x-c)(x-d) \dots \\ &+ \varphi x. (x-a). \frac{\operatorname{d}[(x-b)(x-c)(x-d) \dots ]}{\operatorname{d} x} \\ &+ f x. (x-a)(x-b)(x-c)(x-d) \dots=1+f x. (\upsilon-Y)+\varphi x. \frac{\operatorname{d}(\upsilon-Y)}{\operatorname{d} x} \end{aligned}$$

For brevity, let us express

$$f x. (y-Y)+\varphi x. \frac{\operatorname{d}(y-Y)}{\operatorname{d} x}$$

which is a function of the variables $x,$ $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., by

$$ F (x, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime} \text{ etc.})$$

Then we will have the identities [1]

$$\begin{aligned} \varphi & a. (a-b)(a-c)(a-d) \ldots=1+F (a, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.}) \\ \varphi & b. (b-a)(b-c)(b-d) \ldots=1+F (b, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.}) \\ \varphi & c. (c-a)(c-b)(c-d) \ldots=1+F (c, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})\\ & \text{etc.} \end{aligned}$$

Supposing, therefore, that the product of all

$$\begin{aligned} 1 & + F(a, l', l, l', \text{etc.}) \\ 1 & + F(b, l', l, l', \text{etc.}) \\ 1 & + F(c, l', l, l', \text{etc.}) \\ & \text{etc.} \\ \end{aligned}$$

which will be a complete function of the indeterminates $a, b, c,$ etc., $l', l, l',$  etc., and indeed a symmetric function with respect to the same $a, b, c,$  etc., can be represented by

$$\psi(\lambda', \lambda, \lambda, \text{etc.}, l', l, l, \text{etc.})$$

and by multiplying all the equations [1], the result is a new identity [2]

$$\pi \varphi a \cdot \varphi b \cdot \varphi c \ldots = \psi(\lambda', \lambda, \lambda, \text{etc.}, \lambda', \lambda, \lambda, \text{etc.})$$

Moreover it is clear, since the product $\varphi a. \varphi b. \varphi c \ldots$ involves the indeterminates $a,$  $b,$  $c$  etc. symmetrically, that it is possible to find an integral function of the indeterminates $l',$  $l,$  $l'$  etc., which is transformed into $\varphi a. \varphi b. \varphi c \ldots$ by the substitutions $l' = \lambda',$  $l = \lambda,$  $l = \lambda$  etc.  Letting $t$  be that function, we will have the identity [3]

$$p t = \psi(l', l, l, \text{etc.}, l', l, l, \text{etc.})$$

since the substitutions $l' = \lambda',$ $l = \lambda,$  $l = \lambda,$  etc. transform this equation into the identity [2].

Now, from the very definition of the function $F,$ it follows that

$$F(x, L', L, L', \text{etc.}) = 0$$

identically. Hence we also have identities

$$\begin{aligned} 1 & + F(a, L', L, L', \text{etc.}) = 1 \\ 1 & + F(b, L', L, L', \text{etc.}) = 1 \\ 1 & + F(c, L', L, L', \text{etc.}) = 1 \\ & \text{etc.} \\ \end{aligned}$$

and consequently

$$\psi(\lambda', \lambda, \lambda, \text{etc.}, L', L, L, \text{etc.}) = 1$$

and therefore also [4]

$$\psi(l', l, l, \text{etc.}, L', L, L, \text{etc.}) = 1$$

Wherefore, by the combination of equations [3] and [4], and by substituting $l' = L',$ $l = L,$  $l = L$  etc., we will have [5]

$$P T = 1$$

if we denote by $T$ the value of the function $t$  corresponding to those substitutions. Since this value necessarily becomes a finite quantity, $P$ cannot certainly be $=0.$  Q.E.D.

10.
From the foregoing, it is now clear that any integral function $Y$ of a single indeterminate $x,$  whose determinant is $=0,$  can be decomposed into factors, none of which has determinant $0.$  Indeed, once we have found the highest common divisor of the functions $Y$  and $\frac{\operatorname{d} Y}{\operatorname{d} x},$  it will already be resolved into two factors. If one of these factors again has a determinant of $0,$ it can be resolved into two factors in the same way, and thus we shall continue until $Y$  is finally resolved into factors of such a nature that none of them has a determinant of $0.$

Moreover, it is easy to see that among the factors into which $Y$ is resolved, there must be at least one such that among the prime factors of its order, two occurs no more often than among the factors of the order $m$  of the function $Y{:}$  more precisely, if we assume that $m = k \cdot 2^{\mu},$  where $k$  denotes an odd number, then among the factors of the function $Y$  there will be at least one whose order is $k' \cdot 2^{\nu},$  where $k'$  is also odd, and either $\nu=\mu,$  or $\nu<\mu.$  The truth of this assertion follows automatically from the fact that $m$  is the sum  of all of the orders of all of the factors of ${Y}.$

11.
Before we proceed further, we will exhibit a certain expression, the introduction of which is of great utility in all investigations concerning symmetric functions, and which will also be very convenient for us. Let $M$ be a function of some of the indeterminates $a,$  $b,$  $c$  etc., and let $\mu$  be the number of these which enter into the expression of $M,$  without any consideration of the other indeterminates that $M$  may happen to involve. By permuting these $\mu$ indeterminates in all possible ways, both among themselves and with the remaining $m-\mu$  from $a,$  $b,$  $c,$  etc., other expressions similar to $M$  will arise from $M,$  and altogether there will be

$$m(m-1)(m-2)(m-3) \ldots (m-\mu+1)$$

expressions, including $M$ itself, which we shall simply call the complex of all $M$ . Hence it is clear what is meant by the sum of all $M,$ the product of all $M,$  etc. Thus, for example, $\pi$  will denote the product of all $a-b,$  $\upsilon$  the product of all $x-a,$  $\upsilon'$  the aggregate of all $\frac{\upsilon}{x-a},$  etc.

If it happens that $M$ is a symmetric function with respect to some of the $\mu$  indeterminates it contains, then those permutations do not alter the function $M.$  Consequently, in the complex of all $M,$  each term will be repeated, at least $1.2.3 \dots \nu$  times, where $\nu$  is the number of indeterminates with respect to which $M$  is symmetric. However, if $M$ is not only symmetric with respect to $\nu$  indeterminates, but also with respect to $\nu'$  others, and also with respect to $\nu$  others, etc., then $M$  will not be altered, whether the pairs from the first $\nu$  indeterminates are permuted among themselves, or the pairs from the second $\nu',$  or the pairs from the third $\nu,$  etc., so that

$$1 . 2 . 3 \ldots \nu, \quad 1. 2 . 3 \ldots \nu', \quad 1. 2 . 3 \ldots \nu'' \text{ etc.}$$

permutations always give rise to identical terms. Therefore, if we always retain only one of these identical terms, we will have altogether

$$\frac{m(m-1)(m-2)(m-3) \ldots (m-\mu+1)}{1. 2 . 3 \ldots \nu. 1 . 2 . 3 \ldots \nu'. 1 . 2 . 3 \ldots \nu'' \text{etc.}}$$

terms, which we shall call the complex of all $M$ excluding repetitions, to distinguish it from the complex of all $M$  including repetitions. Whenever nothing is expressly stated to the contrary, repetitions will always be understood to be admitted.

Of course, it is easy to see that the sum of all $M,$ or the product of all $M,$  or generally any symmetric function of all $M$  always becomes a symmetric function of the indeterminates $a,$  $b,$  $c$  etc., whether repetitions are allowed or excluded.

12.
We now let $u,$ $x$  be  indeterminates, and consider the product of all $u-(a+b) x+a b,$  excluding repetitions, which we will denote by $\zeta.$  Thus, $\zeta$  will be the product of $\frac{1}{2} m(m-1)$  factors

$$ \begin{array}{c} u-(a+b) x+a b \\ u-(a+c) x+a c \\ u-(a+d) x+a d \\ \text{ etc. } \\ u-(b+c) x+b c \\ u-(b+d) x+b d \\ \text{ etc. } \\ u-(c+d) x+c d \\ \text{ etc. etc. etc. } \end{array} $$

This function, since it symmetrically implicates the indeterminates $a,$ $b,$  $c,$  etc., can be assigned an entire function of the indeterminates $u,$  $x,$  $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime},$  etc., denoted by $z,$  which will pass into $\zeta$  if the indeterminates l^{\prime}, $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., are replaced by $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime},$  etc. Finally, let us denote by $Z$  the function of the indeterminates $u,$  $x$  alone, which is transformed into $z$  if we assign determinate values $L^{\prime},$  $L^{\prime \prime},$  $L^{\prime \prime \prime},$  etc., to the indeterminates $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc.

These three functions $\zeta,$ $z,$  $Z$  can be considered as integral functions of order $\frac{1}{2} m(m-1)$  of the indeterminate $u,$  with indeterminate coefficients, which

for $\zeta,$ will be functions of the indeterminates $x,$ $a,$ $b,$  $c$  etc. for $z,$  will be functions of the indeterminates $x,$  $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc. for $Z,$  will be functions of the indeterminate $x$  alone.

However, the individual coefficients of $z$ will be transformed into the coefficients of $\zeta$  by substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime},$  etc., and into the coefficients of $Z$  by substitutions $l^{\prime}=L^{\prime},$  $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime},$  etc. The same considerations regarding the coefficients will also hold for the determinants of the functions $\zeta,$  $z,$  $Z.$  And we will now inquire more closely into these, with the aim of demonstrating the following theorem:

Whenever $P \neq 0,$ the determinant of the function $Z$  cannot be identically $=0.$ 

13.
The proof of this theorem would indeed be very easy if we were allowed to assume that $Y$ can be resolved into simple factors

$$(x-A)(x-B)(x-C)(x-D) \ldots$$

Then it would also be certain that $Z$ is the product of all $u-(A+B) x+A B,$  and that the determinant of the function $Z$  is the product of the differences between pairs of the quantities

$$\begin{array}{c} (A+B) x-A B \\ (A+C) x-A C \\ (A+D) x-A D \\ \text{etc.} \\ (B+C) x-B C \\ (B+D) x-B D \\ \text{etc.} \\ (C+D) x-C D \\ \qquad \text{etc. etc.}\\ \end{array}$$

However, this product cannot identically vanish unless one of the factors becomes identically $=0.$ It would then follow that two of the quantities $A,$  $B,$  $C$  etc. would be equal, and consequently, the determinant $P$  of the function $Y$  would be $=0,$  contrary to the hypothesis.

Setting aside such reasoning, which is clearly begging the question (just as in art. 6), we now proceed to a rigorous demonstration of the theorem of art. 12.

14.
The determinant of the function $\zeta$ will be the product of all differences between pairs $(a+b) x-a b,$  the number of which is:

$$\frac{1}{2} m(m-1) \left(\frac{1}{2} m(m-1)-1\right)=\frac{1}{4}(m+1) m(m-1)(m-2)$$

This number represents the order of the determinant of the function $\zeta$ with respect to the indeterminate $x.$  The determinant of the function $z$  will indeed be of the same order. However, the determinant of the function $Z$ may belong to a lower order in the case where the coefficients of some of the highest powers of $x$  vanish. It is now our task to demonstrate that all coefficients in the determinant of the function $Z$ cannot vanish.

Upon closer examination of these differences, whose product is the determinant of the function $\zeta,$ we will find that some of them (namely, the differences between pairs of $(a+b) x-ab$  which have a common element) will provide

the product of all $(a-b)(x-c)$

while others (namely, the differences between pairs $(a+b) x-a b,$ whose elements are different) will result in:

the product of all $(a+b-c-d) x-a b+c d$ excluding repetitions.

Each factor in the former product will appear $m-2$ times, and each factor $x-c$  will appear $(m-1)(m-2)$  times. Therefore, we conclude that this product becomes:

$$=\pi^{m-2} \upsilon^{(m-1)(m-2)}$$

If we denote this latter product by $\rho,$ the determinant of the function $\zeta$  will be:

$$=\pi^{m-2} \upsilon^{(m-1)(m-2)} \rho$$

Furthermore, if we let $r$ denote the function of the indeterminates $x,$  $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which is transformed into $\rho$  by the substitutions $l^{\prime}=\lambda^{\prime},$  $l^{\prime \prime}=\lambda^{\prime \prime},$  $l^{\prime \prime \prime}=\lambda^{\prime \prime \prime}$  etc., and by ${R}$  the function of only $x,$  into which $r$  is transformed by the substitutions $l^{\prime}=L^{\prime},$  $l^{\prime \prime}=L^{\prime \prime},$  $l^{\prime \prime \prime}=L^{\prime \prime \prime}$  etc., it is clear that the determinant of the function $z$  becomes

$$=p^{m-2} y^{(m-1)(m-2)} r$$

and the determinant of the function $Z$ becomes

$$=P^{m-2} Y^{(m-1)(m-2)} R$$

Therefore, since $P$ is not $=0$  by hypothesis, it remains to be demonstrated that $R$  cannot vanish identically.

15.
To this end, we introduce another indeterminate $w,$ and consider the product of all

$$(a+b-c-d) w+(a-c)(a-d)$$

excluding repetitions. Since this product involves $a,$ $b,$  $c$  etc. symmetrically, it can be represented as an integral function of the indeterminates $w,$  $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime},$  etc. We will denote this function as $f (w, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}).$  The number of these factors $(a+b-c-d) w+(a-c)(a-d)$  is

$$\frac{1}{2} m(m-1)(m-2)(m-3)$$

from which we easily deduce that

$$f (0, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})=\pi^{(m-2)(m-3)}$$

and therefore also

$$f (0, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.})=p^{(m-2)(m-3)}$$

and finally

$$f (0, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})=P^{(m-2)(m-3)}$$

Generally speaking, the function $f (w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.})$ will be of order

$$\frac{1}{2} m(m-1)(m-2)(m-3)$$

However, in special cases, it may belong to a lower order, if it happens that some coefficients vanish from the highest power of $w.$ Nevertheless, it is impossible for that function to be identically $=0,$  since the equation just found demonstrates that at least the final term of the function does not vanish. Let us take the highest term of the function $f (w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.}),$ which indeed has a non-vanishing coefficient, to be $N w^{\nu}.$  If we substitute $w=x-a,$  it is clear that $f (x-a, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.})$  is an integral function of the indeterminates $x,$  $a,$  or in other words, it is a function of $x$  with coefficients depending on the indeterminate $a,$  in such a way that the highest term is $N x^{\nu},$  and thus it has a coefficient determined by $a$  that is not $=0.$  Likewise, $f (x-b, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.}),$  $f (x-c, L^{\prime}, L^{\prime \prime}, L^{\prime \prime}, \text{ etc.}),$  etc., will be integral functions of $x,$  whose highest terms are $N x^{\nu},$  while the coefficients of the subsequent terms depend on the indeterminates $a,$  $b,$  $c,$  etc.

Let us now consider the product of the following $m$ factors:

$$\begin{array}{c} f (x-a, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.}) \\ f (x-b, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.}) \\ f (x-c, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.}) \\ \text{etc.} \end{array}$$

Since this product is a function of indeterminates $x,$ $a,$  $b,$  $c,$  etc., $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime},$  etc., and it is symmetric with respect to $a,$  $b,$  $c,$  etc., it can be represented as a function of indeterminates $x,$  $\lambda^{\prime},$  $\lambda^{\prime \prime},$  $\lambda^{\prime \prime \prime},$  etc., $l^{\prime},$  $l^{\prime \prime},$  $l^{\prime \prime \prime}$  etc., which we denote by

$$\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime}, \text{ etc.})$$

Therefore,

$$\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})$$

will be the product of the factors

$$\begin{array}{c} f (x-a, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}) \\ f (x-b, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}) \\ f (x-c, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}) \\ \text{etc.} \end{array}$$

and therefore it will be indefinitely divisible by $\rho,$ as it is easy to see that any factor of $\rho$  is involved in some of these factors. Let us therefore set

$$\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})=\rho \psi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.})$$

where the character $\psi$ represents an integral function. From this it is easily deduced that

$$\varphi (x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})=R \psi (x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})$$

But as we have shown above, the product of factors

$$\begin{array}{c} f (x-a, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}) \\ f (x-b, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}) \\ f (x-c, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}) \\ \text{ etc. } \end{array}$$

which will be $=\varphi (x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}, \text{ etc.}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.})$ will have $N^{m} x^{m v}$  as its highest term; therefore, the same highest term will be present in the function $\varphi (x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime}, \text{ etc.}),$  and hence it cannot be identically $=0.$  Therefore, $R$  cannot be identically $=0,$  and neither can the determinant of the function $Z.$  Q.E.D.

16.
Let $\varphi(u, x)$ denote the product of any number of factors, in which the indeterminates $u, x$ enter linearly only, or equivalently, which are of the form

$$\begin{array}{c} \alpha+\beta u+\gamma x \\ \alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x \\ \alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x \\ \text{ etc. } \end{array}$$

''and let $w$ be another indeterminate. Then the function''

$$\varphi (u+w . \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}, x-w . \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u})=\Omega$$

will be indefinitely divisible by $\varphi(u, x).$ 

Proof. Setting

$$\begin{aligned} \varphi(u, x)= & (\alpha+\beta u+\gamma x) Q \\ = & (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x) Q^{\prime} \\ = & (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x) Q^{\prime \prime} \\ & \text{ etc. } \end{aligned}$$

where $Q, Q^{\prime}, Q^{\prime \prime},$ etc., are integral functions of the indeterminates $u, x, \alpha, \beta, \gamma, \alpha^{\prime}, \beta^{\prime}, \gamma^{\prime},$  $\alpha^{\prime \prime}, \beta^{\prime \prime}, \gamma^{\prime \prime},$  etc., we have

$$\begin{aligned} \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x} & =\gamma Q+(\alpha+\beta u+\gamma x). \frac{\operatorname{d} Q}{\operatorname{d} x} \\ & =\gamma^{\prime} Q^{\prime}+ (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x). \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} x} \\ & =\gamma^{\prime \prime} Q^{\prime \prime}+ (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x). \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} x} \\ & \text{ etc. } \\ \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} & =\beta Q+(\alpha+\beta u+\gamma x). \frac{\operatorname{d} Q}{\operatorname{d} u} \\ & =\beta^{\prime} Q^{\prime}+ (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x). \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} u} \\ & =\beta^{\prime \prime} Q^{\prime \prime}+ (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x). \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} u} \\ & \text{ etc.} \end{aligned}$$

Substituting these values into the factors from which the product $Q$ is formed, namely

$\begin{array}{c} \alpha+\beta u+\gamma x+\beta w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}-\gamma w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} \\ \alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x+\beta^{\prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}-\gamma^{\prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} \\ \alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x+\beta^{\prime \prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} x}-\gamma^{\prime \prime} w. \frac{\operatorname{d} \varphi(u, x)}{\operatorname{d} u} \\ \end{array}$ etc. resp.

yields the following values:

$\begin{array}{c} (\alpha+\beta u+\gamma x) \left(1+\beta w . \frac{\operatorname{d} Q}{\operatorname{d} x}-\gamma w . \frac{\operatorname{d} Q}{\operatorname{d} u}\right) \\ (\alpha^{\prime}+\beta^{\prime} u+\gamma^{\prime} x) \left(1+\beta^{\prime} w . \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} x}-\gamma^{\prime} w . \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} u}\right) \\ (\alpha^{\prime \prime}+\beta^{\prime \prime} u+\gamma^{\prime \prime} x) \left(1+\beta^{\prime \prime} w . \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} x}-\gamma^{\prime \prime} w . \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} u}\right) \\ \end{array}$ etc.|undefined

Therefore, $\Omega$ will be the product of $\varphi(u, x)$  with the factors

$$\begin{aligned} & 1+\beta w. \frac{\operatorname{d} Q}{\operatorname{d} x}-\gamma w. \frac{\operatorname{d} Q}{\operatorname{d} u} \\ & 1+\beta^{\prime} w. \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} x}-\gamma^{\prime} w. \frac{\operatorname{d} Q^{\prime}}{\operatorname{d} u} \\ & 1+\beta^{\prime \prime} w. \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} x}-\gamma^{\prime \prime} w. \frac{\operatorname{d} Q^{\prime \prime}}{\operatorname{d} u} \end{aligned}$$

etc., i.e. of $\varphi(u, x)$ with an integral function of the indeterminates $u,$  $x,$  $w,$  $\alpha,$  $b,$  $\gamma,$  $\alpha^{\prime},$  $b^{\prime},$  $\gamma^{\prime},$  $\alpha^{\prime \prime},$  $b^{\prime \prime},$  $\gamma^{\prime \prime}$  etc. Q.E.D.

17.
The theorem of the previous article clearly applies to the function $\zeta,$ which we suppose to be given by

$$f (u, x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})$$

so that

$$f (u+w . \frac{\operatorname{d} \zeta}{\operatorname{d} x}, x-w . \frac{\operatorname{d} \zeta}{\operatorname{d} u}, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})$$

will be indefinitely divisible by $\zeta.$ We will represent the quotient, which is a function of the indeterminates $u,$  $x,$  $w,$  $a,$  $b,$  $c$  etc., and is symmetric with respect to the same $a,$  $b,$  $c$  etc., by

$$\psi (u, x, w, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime} \text{ etc.})$$

From this, we deduce the identities

$$f (u+w . \frac{\operatorname{d} z}{\operatorname{d} x}, x-w . \frac{\operatorname{d} z}{\operatorname{d} u}, l^{\prime}, l^{\prime \prime}, l^{\prime \prime} \text{ etc.})=z \psi (u, x, w, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime} \text{ etc.})$$

and

$$f (u+w . \frac{\operatorname{d} Z}{\operatorname{d} x}, x-w . \frac{\operatorname{d} Z}{\operatorname{d} u}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime} \text{ etc.})=Z \psi (u, x, w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

Therefore, if we simply represent the function $Z$ as ${F}({u}, x),$  so that we have

$$f (u, x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})=F(u, x)$$

we obtain the identity

$$F (u+w . \frac{\operatorname{d} Z}{\operatorname{d} x}, x-w . \frac{\operatorname{d} Z}{\operatorname{d} u})=Z \psi (u, x, w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

18.
Thus, if for determinate values $u=U,$ $x=X$  of $u,$  $x,$  we obtain

$$\frac{\operatorname{d} Z}{\operatorname{d} x}=X^{\prime}, \quad \frac{\operatorname{d} Z}{\operatorname{d} u}=U^{\prime}$$

we will have an identity

$$F (U+w {X}^{\prime}, {X}-w U^{\prime})={F}(U, {X}). \psi (U, X, w, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

Whenever $U^{\prime}$ does not vanish, it will be permissible to establish

$$w=\frac{X-x}{U^{\prime}}$$

and from this we obtain

$$F (U+\frac{X X^{\prime}}{U^{\prime}}-\frac{X^{\prime} x}{U^{\prime}}, x)=F(U, X). \psi (U, X, \frac{X-x}{U^{\prime}}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

which can also be stated as follows:


 * If we set $u=U+\frac{X X^{\prime}}{U^{\prime}}-\frac{X^{\prime} x}{U^{\prime}}$ in the function $Z,$  it is transformed into

$$F(U, X). \psi (U, X, \frac{X-x}{U^{\prime}}, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})$$

19.
Therefore, since in the case where $P=0$ does not hold, the determinant of the function $Z$  is a function of the indeterminate $x$  which does not vanishing by itself, it is clear that the multitude of determinate values of $x$  through which this determinant can attain the value $0$  will be a finite number, so that infinitely many determinate values of $x$  can be assigned that make this determinant different from $0.$  Let $X$  be such a value of $x$  (which can also be assumed to be real). Then the determinant of the function ${F}(u, X)$ will not be $=0,$  and hence it follows, by theorem II. of art. 6, that the functions

$F(u, X)$ and $ \frac{\operatorname{d} F(u, X)}{\operatorname{d} u}$

cannot have any common divisor. Furthermore, let us suppose that there exists some determinate value of $u,$ say $U$  (whether it be real or imaginary, i.e. of the form $g+h \sqrt{-1}\,$ ), which makes $F(u, X)=0,$  i.e. ${F}(U, X)=0.$  Therefore, $u-U$  will be an indefinite factor of the function ${F}(u, {X}),$  and consequently the function $\frac{\operatorname{d} F(u, X)}{\operatorname{d} u}$  will not be divisible by $u-U.$  Therefore, assuming that this function $\frac{\operatorname{d} F(u, X)}{\operatorname{d} u}$  attains the value $U^{\prime}$  when $u=U,$  it cannot be true that $U^{\prime}=0.$  However, $U^{\prime}$  will clearly be the value of the partial differential quotient $\frac{\operatorname{d} Z}{\operatorname{d} u}$  for $u=U,$  $x={X}.$  Therefore, if we denote the value of the partial differential quotient $\frac{\operatorname{d} Z}{\operatorname{d} x},$  for the same values of $u,$  $x,$  by $X^{\prime},$  then it is evident from what has been demonstrated in the previous article that the function $Z$  will vanish identically upon substituting

$$u=U+\frac{X X^{\prime}}{U^{\prime}}-\frac{X^{\prime} x}{U^{\prime}}$$

and therefore it will be indefinitely divisible by the factor

$$u+\frac{X^{\prime}}{U^{\prime}} x- \left(U+\frac{X X^{\prime}}{U^{\prime}}\right)$$

Consequently, by setting $u=x x,$ it is evident that ${F}(x x, x)$  will be divisible by

$$x x+\frac{X^{\prime}}{U^{\prime}} x- \left(U+\frac{X X^{\prime}}{U^{\prime}}\right)$$

and therefore it will obtain the value $0$ if $x$  is taken to be the root of the equation

$$x x+\frac{X^{\prime}}{U^{\prime}} x- \left(U+\frac{X X^{\prime}}{U^{\prime}}\right)=0$$

i.e. if we set

$$x=\frac{-X^{\prime} \pm \sqrt{(4 U U^{\prime} U^{\prime}+4 X X^{\prime} U^{\prime}+X^{\prime} X^{\prime})}}{2 U^{\prime}}$$

whose values are known to be either real or of the form $g+h \sqrt{-1}.$

It is easy to show that, for the same values of $x,$ the function $Y$  must also vanish. For clearly $f (x x, x, \lambda^{\prime}, \lambda^{\prime \prime}, \lambda^{\prime \prime \prime}\text{ etc.})$ is the product of all $(x-a)(x-b)$  without repetitions, and thus it is $=v^{m-1}.$  Hence it follows automatically that

$$\begin{aligned} & f (x x, x, l^{\prime}, l^{\prime \prime}, l^{\prime \prime \prime} \text{ etc.})=y^{m-1} \\ & f (x x, x, L^{\prime}, L^{\prime \prime}, L^{\prime \prime \prime} \text{ etc.})=Y^{m-1} \end{aligned}$$

or ${F}(x x, x)=Y^{m-1},$ whose determined value therefore cannot vanish unless the value of $Y$  vanishes simultaneously.

20.
By the aid of the preceding investigations, the solution of the equation $Y=0,$ i.e., the finding of a determined value of $x,$  either real or in the form $g+h \sqrt{-1},$  which satisfies it, has been reduced to the solution of the equation ${F}(u, X)=0,$  if indeed the determinant of the function $Y$  is not $=0.$  It is worth noting that if all coefficients in $Y,$  i.e., the numbers $L^{\prime},$  $L^{\prime \prime},$  $L^{\prime \prime \prime},$  etc., are all real quantities, then all coefficients in $F(u, X)$  will also be real, provided that, as is permissible, a real value is taken for $X.$  The order of the secondary equation $F(u, X)=0$  is $\frac{1}{2} m(m-1){:}$  hence, whenever $m$  is a even number of the form $2^{\mu} k,$  where $k$  is an indefinite odd number, the order of the secondary equation will be of the form $2^{\mu-1} k.$

In the case where the determinant of the function $Y$ is $=0,$  we can find (by art. 10) another function $\mathfrak{Y}$  which divides it, whose determinant is not $=0,$  and whose order is of the form $2^{\nu} k,$  with either $\nu<\mu$  or $\nu=\mu.$  Any solution of the equation $\mathfrak{Y}=0$  will also satisfy the equation $Y=0{:}$  the solution of the equation $\mathfrak{Y}=0$  can then be reduced to the solution of another equation, whose order is of the form $2^{\nu-1} k.$

From these considerations, we infer that the general solution of any equation, whose order is an even number of the form $2^{\mu} k,$ can be reduced to the solution of another equation, whose order is of the form $2^{\mu^{\prime}} k,$  such that $\mu^{\prime}<\mu.$  If this number is even, i.e. if $\mu^{\prime}$  is not $=0,$  the same method can be applied again, and so we can continue until we reach an equation whose order is expressed by an odd number; and the coefficients of this equation will all be real, since all coefficients of the original equation were real. Such an equation of odd order is known to be solvable, and indeed has a real root, whence each preceding equation will also be soluble, either by real roots or by roots of the form $g+h \sqrt{-1}.$

It is therefore evident that any function $Y$ of the form $x^{m}-L^{\prime} x^{m-1}+L^{\prime \prime} x^{m-2}-$  etc., where $L^{\prime},$  $L^{\prime \prime}$  etc., are determined real quantities, involves an indefinite factor $x-A,$  where $A$  is a quantity either real or contained in the form $g+h \sqrt{-1}.$  In the latter case, it is easily seen that $Y$  also obtains the value $0$  upon substituting $x=g-h \sqrt{-1},$  thus it is divisible by $x-(\dot{g}-h \sqrt{-1}),$  and therefore also by the product $x x-2 g x+g g+h h.$  Therefore, any function $Y$  certainly involves an indefinite real factor of the first or second order, and since the same applies again to the quotient, it is evident that $Y$  can be resolved into real factors of the first or second order. This demonstrates the proposition of this commentary.