Difference between revisions of "1992 USAMO Problems/Problem 5"
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Hence, for <math>i\in\left\{1;\;2;\;\ldots;\;k\right\}</math>, we have <math>R\left(z_{i}\right)=Q\left(z_{i}\right)\left(Q\left(z_{i}\right)-Q\left(z_{k+1}\right)\right)=0 </math> because <math>Q\left(z_{i}\right)=0,</math> and <math>R\left(z_{k+1}\right)=</math><math>Q\left(z_{k+1}\right)\left(Q\left(z_{k+1}\right)-Q\left(z_{k+1}\right)\right)=0</math>. Thus, <math>R\left(z_{i}\right)=0</math> holds for every <math>i\in\left\{1;\;2;\;\ldots;\;k+1\right\}</math>, and this proves the lemma for <math>n = k + 1</math>. Hence, the induction step is complete, and the lemma is proven. | Hence, for <math>i\in\left\{1;\;2;\;\ldots;\;k\right\}</math>, we have <math>R\left(z_{i}\right)=Q\left(z_{i}\right)\left(Q\left(z_{i}\right)-Q\left(z_{k+1}\right)\right)=0 </math> because <math>Q\left(z_{i}\right)=0,</math> and <math>R\left(z_{k+1}\right)=</math><math>Q\left(z_{k+1}\right)\left(Q\left(z_{k+1}\right)-Q\left(z_{k+1}\right)\right)=0</math>. Thus, <math>R\left(z_{i}\right)=0</math> holds for every <math>i\in\left\{1;\;2;\;\ldots;\;k+1\right\}</math>, and this proves the lemma for <math>n = k + 1</math>. Hence, the induction step is complete, and the lemma is proven. | ||
− | == | + | == See Also == |
{{USAMO box|year=1992|num-b=4|after=Last Question}} | {{USAMO box|year=1992|num-b=4|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] |
Revision as of 07:53, 19 July 2016
Problem 5
Let be a polynomial with complex coefficients which is of degree
and has distinct zeros. Prove that there exist complex numbers
such that
divides the polynomial
.
Solution
Since the zeros of the polynomial
are distinct, the polynomial
divides the polynomial
if and only if
for every
. So it is enough to show that for any
complex numbers
, there exist
complex numbers
, such that the polynomial
satisfies
for every
. This can be generalized:
Lemma) Let be a positive integer, and
be
complex numbers. Then, there exist
complex numbers
such that the polynomial
satisfies
for every
.
Proof: We use induction over .
For , the lemma is trivial, since
, so we can take
, and then the polynomial
clearly satisfies
.
Let be a positive integer. Assume that for
, the lemma is true. For any
complex numbers
, there exist
complex numbers
such that the polynomial
satisfies
for every
.
In fact, after our assumption, there exist complex numbers
such that the polynomial
satisfies
for every
. We want to construct our polynomial
from this polynomial
.
In fact, let for every
, then let
and
. Then,
Hence, for , we have
because
and
. Thus,
holds for every
, and this proves the lemma for
. Hence, the induction step is complete, and the lemma is proven.
See Also
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.