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solutions of xy = y x∗
pahio†
2013-03-22 3:28:51
The equation
xy = y x
(1)
has trivial solutions on the line y = x. For other solutions one has the
Theorem. 1◦ . The only positive solutions of the equation (1) with 1 <
x < y are in a parametric form
1
x = (1+u) u ,
1
y = (1+u) u +1
(2)
where u > 0.
2◦ . The only rational solutions of (1) are
n
1
x = 1+
,
n
y =
1
1+
n
n+1
(3)
where n = 1, 2, 3, . . .
3◦ . Consequently, the only integer solution of (1) is
24 = 16 = 42 .
Proof. 1◦ . Let (x, y) be a solution of (1) with 1 < x < y. Set y = x+δ
(δ > 0). Now
xx+δ = (x+δ)x ,
from which we obtain easily
x =
δ
1+
x
xδ
1
:= (1+u) u ,
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1
where u = xδ . Then
δ
y = x+δ = x 1+
x
1
1
= (1+u) u (1+u) = (1+u) u +1 .
2◦ . The unit fractions u = n1 yield from (2) rational solutions (3). Further, no
irrational value of u cannot make both x and y of (2) rational, since otherwise the
ratio 1+u of the latter numbers would be irrational (cf. rational and irrational).
Accordingly, for other rational solutions than (3), we must consider the values
m
u :=
n
with coprime positive integers m, n where m > 1. Make the antithesis that
n
mm
x = 1+
∈ Q.
n
Because the integers coprime with m form a group with respect to the multiplication modulo m (cf. prime residue classes), the congruence
nz ≡ 1
(mod m)
has a solution z. Thus we may write nz = km + 1 and rewrite the rational
number
km+1
n z
m nz
m m
m k m m1
mm
m
= 1+
= 1+
= 1+
1+
. (4)
1+
n
n
n
n
n
m1
This product form tells that 1+ m
is rational. But the number
n
r
m m1
m m+n
1+
=
n
n
cannot be rational without the coprime integers m+n and n both being mth
powers. If we had n = v m , then by Bernoulli inequality,
(v+1)m > v m +mv = n+m,
i.e. m+n could not be a mth power. The contradictory situation means, by (4),
that the antithesis is wrong. Therefore, the numbers (3) give the only rational
solutions of (1).
Note. The value n = 2 in (3) produces x = 49 , y =
27
94
9 8
27
=
.
4
8
27
8 ,
whence (1) reads
The truth of the equality (5) may also be checked by the calculation
" 1 # 27
" # 49
27
94
27
4
3
9 8
9 2
3
27
3 4
=
=
=
.
=
4
4
2
2
8
2
(5)
References
[1] P. Hohler & P. Gebauer: Kann man ohne Rechner entscheiden, ob eπ
oder π e grösser ist? − Elemente der Mathematik 36 (1981).
[2] J. Sondow & D. Marques: Algebraic and transcendental solutions of
some exponential equations. − Annales Mathematicae et Informaticae 37
(2010); available directly at arXiv.
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