1.
18
1.2.
ALGEBRA
and furthermore for all integers ml, m2 2:: 1 . Suppose that for all integers k 2:: 1 we have I[k](x) iBecause there are n! permutations, it follows that for k such that positive integers nl >
x.
n2
ai-, in=2
Let h = n l since numbers or
(2)
l[n2+h](x)
=
n!
x
E
[a, b],
(3)
= (10 l[n2-1])(x), X [a, b]. Because I is injective, we obtain l[n2+h-l] (x) l[n2-1] (x), x [a, b] (10 l[n2+h-l])(x)
and in the same manner
= I(x), l[h](X) = x, x
l[h+l](X)
or
Alternative solution.
E
E
=
x E
E
[a, b]
[a, b].
Let Sn be the symmetric group of order n and Hn the cyclic subgroup generated by a. It is clear that Hn is a finite group and therefore there is integer h such that a[h) is identical permutation. Notice that if = 1, n l I k ](X) = otherwIse
{ axq[k1(i)
x
�i' i =
= x and the solution is complete. (Dorin Andrica, Revista Matematidi Timi§oara (RMT) , No. 2(1978) , pp. 53,
Then I[h](x) Problem 3540)
x + y + z = 0, we obtain x + y -z, y + z -x, z + x -y, or, by squaring and rearranging, x2 + y2 = Z2 2xy, y2 + Z2 = x2 2yz, Z2 + x2 y2 - 2zx. The given equality is equivalent to x2 - 2yz y2 - 2zx x3 y3 Z3 Z2 - 2xy -z--=- + -x + -y -yz + -zx + -, xy 4.
Because
=
_
--
=
=
_
=
=
Z3 -(x +y + z) + 2 ( -Xyz + -yzx + -zxY ) = yz-x3 + zx-y3 + -. xy
2(X2 y2 + y2z2 + 2 X2 ) = x4 + y4 + Z4 . On the other side, from x + y + z = 0 we obtain (x + y + Z) 2 = 0 or x2 + y2 + Z2 = -2( xy + yz + zx). Squaring yields X4 + y4 + Z4 + 2(X2y2 + y2z2 + Z2 X2 ) 4(X2 y2 + y2z 2 + Z2X2 ) + 8xyz(x + y + z) Z
(3)
l[n2](x),
and consequently to
The last equality is equivalent to >
0 and observe that for all k the functions I[k] are injective, 1 , n are distinct. From relation we derive that
>
19
SOLUTIONS
=
or as desired.
(Titu Andreescu,
Revista Matematica Timi§oara (RMT) , No. 3(1971), pp. 25, Problem 483; Gazeta Matematica (GM-B) , No. 12(1977) , pp. 501, Problem 6090)
a, b,
d are different from zero. Consider the equation x4 - (L a) x3 + (L ab) x2 - (L abc) x + abed 0 with roots a, b, e, d. Substituting x with a, b, e and d and simplifying by a, b, e, d i- 0, 5. We assume that numbers
c,
=
after summing up we obtain
Because
If
L
a=
0, it follows that
L a3 = 3 L abe.
one of the numbers is zero, say
a,
then
b+ e + d = 0, or b e = -d. It is left to prove that b3 e3 + d3 = 3bed. Now b3 + e3 + d3 = b3 + e3 - (b + e) 3 -3be(b + ) = 3bed as desired. (Dorin Andrica, Revista Matematica Timi§oara (RMT) , No. 1-2(1979) , pp. 47, +
+
=
Problem 3803)
6. Because
a+b+e
a3 + b3 + e3 3abe =
=
a5 + b5 + e5 5abe( a2 + b2 + e2 + ab + be + ea)
0, we obtain
and
c
=