Algebra 429(a 11 +a 12 +a 13 )x 1 +(a 12 +a 13 )m+a 13 n= 0 ,
(a 31 +a 32 +a 33 )x 1 +(a 32 +a 33 )m+a 33 n= 0.The hypothesesa 31 +a 32 +a 33 >0 anda 31 <0 implya 32 +a 33 ≥0, and therefore
(a 32 +a 33 )m≥0 anda 33 n≥0. We deduce thatx 1 ≤0, which combined witha 12 <0,
a 13 <0,a 11 +a 12 +a 13 >0 gives
(a 11 +a 12 +a 13 )x 1 ≤ 0 ,(a 12 +a 13 )m≤ 0 ,a 13 n≤ 0.The sum of these three nonpositive terms can be zero only when they are all zero. Hence
x 1 =0,m=0,n=0, which contradicts our assumption. We conclude that the system
has the unique solutionx 1 =x 2 =x 3 =0.
Second solution: Suppose there is a nontrival solution(x 1 ,x 2 ,x 3 ). Without loss of
generality, we may assume that|x 3 |≥|x 2 |≥|x 1 |. We havea 31 ,a 32 <0 and 0<
−a 31 −a 32 <a 33 ,so
|a 33 x 3 |=|−a 31 x 1 −a 32 x 2 |≤(−a 31 −a 32 )|x 2 |≤(−a 31 −a 32 )|x 3 |<a 33 |x 3 |.This is a contradiction, which proves that the system has no nontrivial solution.
(7th International Mathematical Olympiad, 1965, proposed by Poland)
235.First solution: The zeros ofP(x)are, ^2 ,...,n, whereis a primitive(n+ 1 )st
root of unity. As such, the zeros ofP(x)are distinct. Let
P(xn+^1 )=Q(x)·P(x)+R(x),whereR(x)=an− 1 xn−^1 +···+a 1 x+a 0 is the remainder. Replacingxsuccessively by
, ^2 ,...,n, we obtain
ann−^1 +···+a 1 +a 0 =n+ 1 ,
an(^2 )n−^1 +···+a 1 ^2 +a 0 =n+ 1 ,
...
an(n)n−^1 +···+a 1 n+a 0 =n+ 1 ,or
[a 0 −(n+ 1 )]+a 1 +···+an− 1 n−^1 = 0 ,
[a 0 −(n+ 1 )]+a 1 (^2 )+···+an− 1 (^2 )n−^1 = 0 ,
···
[a 0 −(n+ 1 )]+a 1 (n)+···+an− 1 (n)n−^1 = 0.This can be interpreted as a homogeneous system in the unknownsa 0 −(n+ 1 ),
a 1 ,a 2 ,...,an− 1. The determinant of the coefficient matrix is Vandermonde, thus nonzero,