Difference between revisions of "1976 IMO Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | {{ | + | I shall prove by induction that <math>P_n(x)</math> has <math>2^n</math> distinct real solutions, where <math>2^{n-1}</math> are positive and <math>2^{n-1}</math> are negative. Also, for ever root <math>r</math>, <math>|r|<2</math>. |
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| + | Clearly, <math>P_1(x)</math> has 2 real solutions, where 1 is positive and 1 is negative. The absolute values of these two solutions are also both less than 2. This proves the base case. | ||
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| + | Now assume that for some positive integer <math>k</math>, <math>P_k(x)</math> has <math>2^k</math> distinct real solutions with absolute values less than 2, where <math>2^{k-1}</math> are positive and <math>2^{k-1}</math> are negative. | ||
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| + | Choose a root <math>r</math> of <math>P_{k+1}(x)</math>. Let <math>P_1(r)=s</math>, where <math>s</math> is a real root of <math>P_k(x)</math>. We have that <math>-2<s<2</math>, so <math>0<r^2<4</math>, so <math>r</math> is real and <math>|r|<2</math>. Therefore all of the roots of <math>P_{k+1}</math> are real and have absolute values less than 2. | ||
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| + | Note that the function <math>P_{k+1}(x)</math> is an even function, since <math>P_1(x)</math> is an even function. Therefore half of the roots of <math>P_{k+1}</math> are positive, and half are negative. | ||
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| + | Now assume for the sake of contradiction that <math>P_{k+1}(x)</math> has a double root <math>r</math>. Let <math>P_1(r)=s</math>. Then there exists exactly one real number <math>r</math> such that <math>r^2-2=s</math>. The only way that this could happen is when <math>s+2=0</math>, or <math>s=-2</math>. However, <math>|s|<2</math> from our inductive hypothesis, so this is a contradiction. Therefore <math>P_{k+1}(x)</math> has no double roots. This proves that that the roots of <math>P_{k+1}(x)</math> are distinct. | ||
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| + | This completes the inductive step, which completes the inductive proof. | ||
| + | |||
== See also == | == See also == | ||
{{IMO box|year=1976|num-b=1|num-a=3}} | {{IMO box|year=1976|num-b=1|num-a=3}} | ||
Revision as of 11:59, 10 February 2012
Problem
Let 
and 
for 
Prove that for any positive integer n the roots of the equation 
are all real and distinct.
Solution
I shall prove by induction that 
has 
distinct real solutions, where 
are positive and are negative. Also, for ever root 
, 
.
Clearly, 
has 2 real solutions, where 1 is positive and 1 is negative. The absolute values of these two solutions are also both less than 2. This proves the base case.
Now assume that for some positive integer 
, 
has 
distinct real solutions with absolute values less than 2, where 
are positive and are negative.
Choose a root of 
. Let 
, where 
is a real root of . We have that 
, so 
, so is real and . Therefore all of the roots of 
are real and have absolute values less than 2.
Note that the function is an even function, since is an even function. Therefore half of the roots of are positive, and half are negative.
Now assume for the sake of contradiction that has a double root . Let . Then there exists exactly one real number such that 
. The only way that this could happen is when 
, or 
. However, 
from our inductive hypothesis, so this is a contradiction. Therefore has no double roots. This proves that that the roots of are distinct.
This completes the inductive step, which completes the inductive proof.
See also
| 1976 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
