A theorem of arithmetic and its proof

8 Pages·1849·0.07 MB·English

by  Euler L.

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A theorem of arithmetic and its proof ∗
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b
e Leonhard Euler†
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O The theorem which Ihave set out to stateandprove, Ihad communicated
H long ago throughletters with friends. In these, it was not elegant enough and
. was seen to be altogether in need of attention, especially, with the proof of
h
it not at all obvious, and perhaps searched for mostly in vain. Consequently,
t
a
I express the theorem in this way:
m
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1 If however many numbers a,b,c,d, etc. are given which are be-
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tween themselves unequal, and for each of them fractions are
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2 formed for which the common numerator is unity and the de-
4 nominator indeed for each is the product of all the differences of
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0 this number and the remaining ones, such that these fractions
5 are:
0
/ 1 1
h
, ,
at (a−b)(a−c)(a−d) etc. (b−a)(b−c)(b−d) etc.
m
1
,
:
v (c−a)(c−b)(c−d) etc.
i
X
then the sum of all these fractions will always be equal to zero.
r
a
∗Originally published as Theorema arithmeticum eiusque demonstratio, Commenta-
tiones arithmeticae 2 (1849), 588-592, and reprinted in Leonhard Euler, Opera Omnia,
Series 1: Opera mathematica, Volume 6, Birkh¨auser, 1992. A copy of the original text is
availableelectronicallyatthe EulerArchive,atwww.eulerarchive.org. This paper is E794
in the Enestr¨om index.
†Date of translation: February 18, 2005. Translated from the Latin by Jordan Bell,
2nd year undergraduate in Honours Mathematics, School of Mathematics and Statistics,
Carleton University, Ottawa, Ontario, Canada. Email: [email protected]
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Therefore, for example, if the given numbers are 2, 5, 7, 8, the four
fractions formed from these are:
1 1 1 1
, , , ,
−3·−5·−6 3·−2·−3 5·2·−1 6·3·1
which are reduced to these:
1 1 1 1
− ,+ ,− ,+ ,
3·5·6 3·2·3 5·2·1 6·3·1
and it will be, by effect of the theorem,
1 1 1 1
− + − + = 0.
90 18 10 18
So that there is no trouble from negative signs, the formation of these
fractions can be ordered such that the given numbers are arranged in the
order of their magnitude, either increasing or decreasing. According to this,
for each one the differences of it and each of the remaining ones is taken in
turns and this is taken as the denominator, with the numerator being unity,
the fractions then will be made with the signs + and - assigned alternately.
For example, if the given numbers are
3,8,12,15,17,18,
from each of these the denominators are assembled in this way:
from 3 5· 9· 12· 14· 15· = 113400
8 5· 4· 7· 9· 10· = 12600
12 9· 4· 3· 5· 6· = 3240
15 12· 7· 3· 2· 3· = 1512
17 14· 9· 5· 2· 1· = 1260
18 15· 10· 6· 3· 1· = 2700
and it will be:
1 1 1 1 1 1
− + − + − = 0,
113400 12600 3240 1512 1260 2700
or if each of these are multiplied by 36,
1 1 1 1 1 1
− + − + − = 0,
3150 350 90 42 35 75
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which, if all the fractions are brought to the same denominator 3150, then
by 1−9+35−75+90−42 = 0 it is clear.
3150
Certainlyinthecasewheretwonumbersaregiven, aproofbythetheorem
is not needed, with it clear to be that
1 1
+ = 0;
a−b b−a
while for the case of three numbers a,b,c it is now more recondite, neverthe-
less it is immediately clear to be that
1 1 1
+ + = 0;
(a−b)(a−c) (b−a)(b−c) (c−a)(c−b)
but with more numbers, and indeed in general for however great a number of
them there are, it is hardly enjoyable to discover the truth of this for them
as for these simpler cases.
The truth of this theorem, furthermore extended very widely, may be
shown in the following way:
General theorem
If however many unequal numbers a,b,c,d,e,f, etc. are given, where the
number of them is equal to m, and the differences of each of them with the
remaining ones are formed into the following products:
(a−b)(a−c)(a−d)(a−e)(a−f) etc. = A,
(b−a)(b−c)(b−d)(b−e)(b−f) etc. = B,
(c−a)(c−b)(c−d)(c−e)(c−f) etc. = C,
(d−a)(d−b)(d−c)(d−e)(d−f) etc. = D,
(e−a)(e−b)(e−c)(e−d)(e−f) etc. = E,
(f −a)(f −b)(f −c)(f −d)(f −e) etc. = F,
where the number of factors of each stays constant at m−1, then it will be
not only that
1 1 1 1 1
+ + + + + etc. = 0,
A B C D E
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but even this general rule:
an bn cn dn en
+ + + + = 0,
A B C D E
providing that the exponent n is a positive integral number less than m−1.
Therefore in the example conveyed previously, for which the given num-
bers are 3, 8, 12, 15, 17, 18, not only is it that:
1 1 1 1 1 1
− + − + − = 0,
113400 12600 3240 1512 1260 2700
but moreover for these following fractions it is also true:
3 8 12 15 17 18
− + − + − = 0,
13400 12600 3250 1512 1260 2700
32 82 122 152 172 182
− + − + − = 0,
113400 12600 3240 1512 1260 2700
33 83 123 153 173 183
− + − + − = 0,
113400 12600 3240 1512 1260 2700
34 84 124 154 174 184
− + − + − = 0,
113400 12600 3240 1512 1260 2700
while it is not permitted to procede to higher powers, with each of the de-
nominators formed by five factors.
Proof of the theorem
xn
Thistheoremoriginatedbyconsiderationoftheexpression ,
(x−a)(x−b)(x−c)(x−d) etc.
whichitisknownthat, whenever theexponent nisapositive integralnumber
less than the number of factors inn the denominator, can always be resolved
into these simple fractions:
A B C D
+ + + + etc.,
x−a x−b x−c x−d
for which the denominators themselves are factors of that denominator and
the numerators are constant quantities which indeed do not depend on x,
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each of which may be defined in the following way. With the given expression
equal to these simple fractions, by multiplying by x−a we will have:
xn B(x−a) C(x−a) D(x−a)
= A+ + + + etc.,
(x−b)(x−c)(x−d) etc. x−b x−c x−d
the equality of which remains true for whatever value is taken for x, since
the letters A,B,C,D, etc. do not depend on x. Therefore this equation will
be true if it is taken x = a, from which it will be
an
= A,
(a−b)(a−c)(a−d) etc.
and the value of A itself may thus be known, and it a similar way it is
observed to be
bn cn
B = ,C = ,
(b−a)(b−c)(b−d) etc. (c−a)(c−b)(c−d) etc.
and thus for the others. According to this, by moving the simple fractions
to the other side it will be
xn A B C D
+ + + + + etc. = 0,
(x−a)(x−b)(x−c)(x−d) etc. a−x b−x c−x d−x
and we will necessarily have, considering the number x as if the last of these
numbers a,b,c,d,...,x,
an bn
+ +
(a−b)(a−c)(a−d)...(a−x) (b−a)(b−c)(b−d)...(b−x)
cn xn
+...+ = 0,
(c−a)(c−b)(c−d)...(c−x) (x−a)(x−b)(x−c)...(x−ν)
where ν denotes the second last of the numbers.
This is proof of the given theorem, which nevertheless is not obvious. By
this the truth between the common terms, for which an account be observed
withoutdifficulty, canbeseen, unlessbychanceanothereasierproofisableto
be found. However, by this a rule may not apprehended unless n is a positive
integral number less than the number of factors in each of the denominators,
because the theorem is not in truth consistent then.
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Therefore with n taken as a larger number, the sum of these fractions
no longer vanishes, for which reason we develop this theorem, by means of
which we will be able to deduce the value of the sum of these in all cases,
with, of course, the number of factors set as m−1 and hence the number of
all those given as a,b,c,d,...x equal to m. If it is n = m−1, or n = m or
n > m, then with the fraction
xn
,
(x−a)(x−b)(x−c)(x−d) etc.
the proof ought to be taken to be false. It is then necessary for the sum of
the fractions that are formed to be equal to the integral side, made from the
integral parts added together.
Hence in the case in which n = m−1, where the integral part is unity, the
sum of these fractions is then equal to 1. Therefore in the example discussed
earlier, where for the following example the signs must be changed, it will
be:
185 175 155 125 85 35
− + − + − = 1.
2700 1260 1512 3240 12600 113400
But if it is n = m, the integral part, which is taken from these fractions,
is x+a+b+c+d+ etc., that is the sum of all the given numbers. Therefore,
with the sum of the numbers given in the above example equal to 73, it will
be:
186 176 156 126 86 36
− + − + − = 73.
2700 1260 1512 3240 12600 113400
At thispoint, thewayto findfurther sumscanbeeasily inferred. Namely,
the sum of the given numbers a,b,c,d,...,x is taken, which is equal to P,
then the sum of the products by two, which is equal to Q, and then the sum
of the products by three, which is equal to R, similarly by four, equal to S,
by five, equal to T, and so on, which forms the series:
1+A+B+C+D+ etc., so that it is,
A = P,B = AP −Q,C = BP −AQ+R,D = CP −BQ+AR−S, etc.,
and so,
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case the sum of our fractions will be
n = m−1 1
n = m A = P
n = m+1 B = P2 −Q
n = m+2 C = P3 −2PQ+R
n = m+3 D = P4 −3P2Q+2PR+Q2 −S
n = m+4 E = P5 −4P3Q+3P2R+3PQ2 −2PS −2QR+T
etc.
Or if the sum of the numbers is taken as P, the sum of the squares as Q,
the sum of the cubes as R, the sum of the fourth powers as S, the sum of
the fifth powers as T, etc., it will be as follows:
1 1 1 1 1
A = B,B = P2 + Q,C = P3 + PQ+ R,
2 2 6 2 3
1 1 1 1 1
D = P4 + B2Q+ Q2 + PR+ S, etc.,
24 4 8 3 4
whose values procede by this rule, so that it is:
A = P
1
B = (PA+Q)
2
1
C = (PB+QA+R)
3
1
D = (PC+QB+RA+S)
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etc.
The truth of our theorem has been established on this, not at all arbi-
trarily, if I will have accurately investigated the nature of the expressions on
which the theorem is based. Therefore, if however many numbers a,b,c,d,
etc. are given, by this the expression will then be (a−b)(a−c)(a−d) etc.,
producedby thedifferences ofeach ofthem witheach oftheothers multiplied
together. Therefore if the number of numbers given is equal to n, and z is
taken as an indefinite quantity, from this is formed this product:
(z −a)(z −b)(z −c)(z −d)(z −e) etc.,
which when expanded by multiplication yields
zn −Azn−1 +Bzn−2 −Czn−3 +Dzn−4 − etc.
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By dividing by z −a we will therefore have
zn −Azn−1 +Bzn−2 −Czn−3 + etc.
(z −b)(z −c)(z −d) etc. =
z −a
But if we now set z = a, it is seen to be of the same form (a−b)(a−c)(a−d)
etc. as I had indicated before for the letter A. Then in fact the value of the
numerator as the denominator goes to zero will be:
nan−1 −(n−1)Aan−2 +(n−2)Ban−3 −(n−3)Can−4 + etc.,
which would be
an −Aan−1 +Ban−2 −Can−3 +Dan−4 − etc. = 0,
(conclusion missing)
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