An algebraic problem of finding four numbers, given the products of each of the numbers with the sum of the three others

8 Pages·1862·0.071 MB·English

by  Euler L.

An algebraic problem of finding four numbers,
5 given the products of each of the numbers with
0
0
the sum of the three others
2 ∗

r
p
A
Leonhard Euler
8 †

1
v

2 If the four numbers that are to be searched for are given as v,x,y,z, the
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following four equations will be considered:
1
4
0 v(x+y +z) = a
5
0 x(v +y +z) = b
/
O y(v+x+z) = c
H
z(v +x+y) = d
.
h
t From these equations, with basic methods three unknowns could be succes-
a
m sively eliminated, and the fourth would be able to be dealt with according
: to the resolution of an equation. However, there is no reason why we would
v
i choose one as preferrable to the others to be ultimately determined. Indeed,
X
it is suitable for none of these equations to be the final one determined, but
r
a in fact for a new unknown to be introduced that relates to each equally and
by which the unknowns would be able to be defined. Therefore we sum the
∗OriginallypublishedasProblemaalgebraicumdeinveniendisquatuornumeris,exdatis
totidem productis uniuscuiusque horum numerorum in summas trium reliquorum, Opera
Posthuma 1 (1862), 282–287, and reprinted in Leonhard Euler, Opera Omnia, Series 1:
Opera mathematica, Volume 6, Birkh¨auser, 1992. A copy of the original text is available
electronically at the Euler Archive, at http://www.eulerarchive.org. This paper is E808
in the Enestr¨om index.
†Dateoftranslation: April8,2005. TranslatedfromtheLatinbyJordanBell,2ndyear
undergraduate in Honours Mathematics, School of Mathematics and Statistics, Carleton
University, Ottawa, Ontario, Canada. Email: [email protected]
1
numbers that are to be found in this formula,
v +x+y +z = 2t,
and so then the above equations are changed into these:
v(2t v) = a = 2tv v2, from which v = t √(tt a)
− − − −
x(2t x) = b = 2tx x2 x = t √(tt b)
− − − −
y(2t y) = c = 2ty y2 y = t √(tt c)
− − − −
z(2t z) = d = 2tz z2 z = t √(tt d).
− − − −
We have the beginning of a solution now, such that from the single quantity
t we have the power to determine all four numbers that are being searched
for, whereby all that remains is for us to investigate this quantity t. With
respect to the equation
v +x+y +z = 2t,
by substituting in place of v,x,y,z, the values formed with t it will be:
4t √(tt a) √(tt b) √(tt c) √(tt d) = 2t,
− − − − − − − −
from which this equation follows:
2t = √(tt a)+√(tt b)+√(tt c)+√(tt d),
− − − −
which indeed could be worked through by the method of Newton for ratio-
nalization, but this would be a most tiresome labor. Therefore we will try
resolving this equation with another method.
We put √(tt a) = p, it will be v = t p; in a similar way
− −
√(tt b) = q, x = t q
− −
√(tt c) = r, y = t r
− −
√(tt d) = s, z = t s
− −
and it will be p+q +r +s = 2t; because of the irrationals p,q,r and s,
this equation should be changed into another, in which the letters p,q,r and
s correspond to powers of exponentials, from which by taking the place of
the letters p,q,r and s that are to be made, a rational equation comes into
existence, by which the value of the unknown t can be defined.
2
To this end, we form this equation:
X4 AX3 +BX2 CX +D = 0,
− −
whose four roots are given by the quantities a,b,c,d. Therefore it will be by
the nature of equations:
A = a+b+c+d
B = ab+ac+ad+bc+bd+cd
C = abc+abd+acd+bcd
D = abcd.
Now it is put Y = tt X, that is X = Y +tt, so that we will have made
− −
by substituting this into the former equation:
Y4 4ttY3 +6t4Y2 4t6Y +t8
− − 
+AY3 3AttY2 +3At4Y At6

− − 
+BY2 2BttY +Bt4  = 0.

−
+CY Ctt
−
+D 



The four roots of this equation of Y will be:
tt a,tt b,tt c,tt d.
− − − −
In place of this equation, we put for the sake of brevity this:
Y4 PY3 +QY2 RS +S = 0,
− −
so that it will be:
R = 4tt A
−
Q = 6t4 3Att+B
−
R = 4t6 3At4 +2Btt C
− −
S = t8 At6 +Bt4 Ctt+D.
− −
Next it will be Y = Z2, that is Z = √Y, so that we will have:
±
Z8 PZ6 +QZ4 RZ2 +S = 0.
− −
3
The following will be the eight roots of this equation:
+√(tt a) = +p √(tt a) = p
− − − −
+√(tt b) = +q √(tt b) = q
− − − −
+√(tt c) = +r √(tt c) = r
− − − −
+√(tt d) = +s √(tt d) = s.
− − − −
This equation in eight dimensions is resolved into two biquadratics where for
onetherootsare+p,+q,+r,+s,whilefortheothertherootsare p, q, r, s,
− − − −
which are:
Z4 αZ3 +βZ2 γZ +dδ = 0
− −
Z4 +αZ3 +βZ2 +γZ +δ = 0,
for which it will be by the nature of equations:
α = p+q +r +s
β = pq +pr+ps+qr +qs+rs
γ = pqr +pqs+prs+qrs
δ = pqrs.
Since the product of these two biquadratic equations should be equal to the
former equation of eight dimensions, it will be:
P = α2 2β
−
Q = β2 2αγ +2δ
−
R = γ2 2βδ
−
S = δ2.
Also with it α = p + q + r + s, it will be α = 2t, and then α2 = 4tt, from
which it will be:
α2 2β = 4tt 2β = P = 4tt A,
− − −
therefore β = A. The second equation Q = β2 2αγ +2δ gives:
2 −
A2
6t4 3Att+B = 4γt+2δ,
− 4 −
4
that is,
3 A2 B
δ = 3t4 Att+2γt + .
− 2 − 8 2
Certainly the third equation R = γ2 2βδ will yield
−
4t6 3At4 +2Btt C = γ2 Aδ,
− − −
that is,
Aδ = 4t6 +3At4 2Btt+c+γ2,
− −
then with the above it will be:
4t6 3A2tt+2Aγt A3
− 2 − 8 
+2Btt+ AB = γ2.
2 
C
− 
By having extracted the square root, it will be obtained that,
1 1 1
γ = At √(4t6 +(2B A2)tt A3 + AB C)
± − 2 − 8 2 −
and then
1 1 1 1 1 1
δ = 3t4+ Att A2 + B 2t√(4t6+(2B A2)tt A3 + AB C),
2 − 8 2 ± − 2 − 8 2 −
of which the square will be
25t8+3At6+−1125BAt24t4+−52A85AB43Cttttt−t++186A114B2AB24  ±±2±A121AA2t2t53t  √(4t6+(2B− 12A2)tt− 18A3+ 21AB−C)
− 4  ±2Bt 
which should be equal to S itself, that is this expression: t8 At6 +Bt4
Ctt+D, from which results this equation: − −
24t8+4At6 5A2t4 5A3tt+ 1 A4 +12t5
+−102Bt4+−25A8−B3Cttt−t+186A414B2B2 = −+122AA2t3t  √(4t6+(2B− 21A2)tt− 18A3+ 21AB−C),
−D  +2Bt 
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which reduced to rationals gives:
+ 3 A7
+ 3 A6 256  + 1 A8
16  9 A5B 4096 
27A4B −64  1 A6B
+5A5 −16  + 5 A4C  −256 
4  +7A3C  32  + 3 A4B2 
+3A4 8A3B 4  + 9 A3B2  128 
 −  +9A2B2  16  1 A4D 
12A2B +7A2C  2  +5A3D  −32 
− t8 t6 +5A2D t4 4 tt 1 A2B3  = 0
+24AC  +12AB2   5A2BC  −16 
7ABC −4 +1A2BD
48D 8AD − 3AB3 4
−  −  3B3  −4  + 1 B4 
 12BC  −  5ABD  16 
−  +9C2  −  1B2D 
  +5B2C  −2 
20BC  2  +D2 
−  +6CD  
  


It put for the sake of brevity E = 1A2 B and u = 2t, and it will be:
4 −
+2A3E +9A2E2
  +12AE3
+3A2E +4A2C +28ACE  +4E4
   +80ADE 
+6AC u8 +12AE2 u6 +80DE u4 uu 32DE2 = 0.
   +96CD  − 
12D 8AD +36C2 +64D2
− − +40CE2
 +12CE  +12E3   
  
 
 
This equation then has four positive roots and the same number of negative
roots, which are equal to each other; thus the resolution of the equation can
be accomplished by a biquadratic equation. Moreover, A,B,C,D and E are
known quantities determined by the given a,b,c,d, with it of course
A = a+b+c+d
B = ab+ac+ad+bc+bd+cd
C = abc+abd+acd+bcd
D = abcd
and likewise
1
E = A2 B.
4 −
With this having been discovered indeed, whatever the value of u the quan-
tities being searched for are:
u √(uu 4a) u √(uu 4b)
v = − − ,x = − −
2 2
u √(uu 4c) u √(uu 4d)
y = − − ,z = − − .
2 2
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Another solution
Furthermore, this problem can be solved in another way: From the first
equations it is
a b = (v x)(y +z)b c = (x y)(v+z)
− − − −
a c = (v y)9x+z)b d = (x z)(v +y)
− − − −
a d = (v z)(x+y)c d = (y z)(v +x)
− − − −
from the first and last of the equations is we obtain,
a b c d
v x = − ,v +x = − .
− y +z y z
−
Were it a+b c d = h it will be h = vx yz, and by making it a+b+c+d = k it
− −
2 − 2
will be
k = vx+vy +vz +xy +xz +yz,
therefore
k h = 2yz +(v +x)(y +z),
−
that is, k h = 2yz (c−d)(y+z) = c + d, therefore 2yz = 2dy−2cz, i.e.
− − y z y z
yyz yzz = dy cz; with it s−et yz = t this equations changes into−this:
− −
(d t)y (c t)z = 0.
− − −
It is not taken dy cz = u, and it will be
−
(c t)u (d t)u
y = − and z = − ,
(c d)t (c d)t
− −
for which substituting t = yz in the values yields
(c t)(d t)uu
t = − − ,
(c d)2tt
−
from which comes
(c d)t√t
u = − .
√(cd (c+d)t+tt)
−
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By substituting this into the values of y and z formed above we will have:
(c d)√t
I. y = − and
√(cd (c+d)t+tt)
−
(d t)√t
II. z = − ;
√(cd (c+d)t+tt)
−
and then by adding and subtracting,
(c+d 2t)√t (c d)√t
y +z = − and y z − .
√(cd (c+d)t+tt) − √(cd (c+d)t+tt)
− −
Then as formerly it is deduced,
√(cd (c+d)t+tt) (a b)√(cd (c+d)t+tt)
v +x = − and v x = − −
√t − (c+d 2t)√t
−
from which, by once more adding and subtracting, with it put for the sake
of previty
b+c+d a = m and a+c+d b = n,
− −
the following values are produced for v and x:
(n 2t)√(cd (c+d)t+tt)
III. v = − − and
2(c+d 2t)√t
−
(m 2t)√(cd (c+d)t+tt)
IV. x = − − .
2(c+d 2t)√t
−
With what we have discovered above and also that h = vx yz, by now
−
substituting in the values of v,x,y,z and unravelling their expansion for the
purpose of determining the value of t, this equation of four dimensions is
formed:
4t(h+t)(c+d 2t)2 = (m 2t)(n 2t)(c t)(d t).
− − − − −
8