Difference between revisions of "2007 AIME II Problems/Problem 14"
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The answer is clearly correct, but the proof has a gap, i.e. there is no reason that <math>f(-2)\neq1</math>. Since <math>f(x)</math> has no real roots, the degree must be even. Consider <math>g(x)= f(x)/f(-x)</math>. Then since <math>f</math> is non-zero, <math>g(x)=g(2x^3+x)</math>. Now the function <math>2x^3+x</math> applied repeatedly from some real starting value of x becomes arbitrarily large, and the limit of <math>g(x)</math> as <math>|x|</math> approaches infinity is 1, so <math>g(x)</math>=1 for all x, or <math>f(x)=f(-x)</math>. Then <math>f(x)=h(x^2+1)</math> for some polynomial <math>h(x)</math>, and <math>h(x^2+1)h(4x^4+1)=h(4x^6+4x^4+x^2+1) = h((x^2+1)(4x^4+1))</math>. Now suppose h has degree m. It is clearly monic. Assume that the next highest non-zero coefficient in h is k. Then, subtracting <math>((x^2+1)(4x^4+1))^m</math> from both sides of the equation yields a polynomial equality with degree <math>4m+2k</math> on the left and degree <math>6k</math> on the right, a contradiction. So <math>h(x)=x^m</math>, and <math>f(x)=(1+x^2)^m</math>. | The answer is clearly correct, but the proof has a gap, i.e. there is no reason that <math>f(-2)\neq1</math>. Since <math>f(x)</math> has no real roots, the degree must be even. Consider <math>g(x)= f(x)/f(-x)</math>. Then since <math>f</math> is non-zero, <math>g(x)=g(2x^3+x)</math>. Now the function <math>2x^3+x</math> applied repeatedly from some real starting value of x becomes arbitrarily large, and the limit of <math>g(x)</math> as <math>|x|</math> approaches infinity is 1, so <math>g(x)</math>=1 for all x, or <math>f(x)=f(-x)</math>. Then <math>f(x)=h(x^2+1)</math> for some polynomial <math>h(x)</math>, and <math>h(x^2+1)h(4x^4+1)=h(4x^6+4x^4+x^2+1) = h((x^2+1)(4x^4+1))</math>. Now suppose h has degree m. It is clearly monic. Assume that the next highest non-zero coefficient in h is k. Then, subtracting <math>((x^2+1)(4x^4+1))^m</math> from both sides of the equation yields a polynomial equality with degree <math>4m+2k</math> on the left and degree <math>6k</math> on the right, a contradiction. So <math>h(x)=x^m</math>, and <math>f(x)=(1+x^2)^m</math>. | ||
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== See also == | == See also == |
Latest revision as of 00:55, 26 November 2020
Problem
Let be a polynomial with real coefficients such that and for all , Find
Solution
Let be a root of . Then we have ; since is a root, we have ; therefore is also a root. Thus, if is real and non-zero, , so has infinitely many roots. Since is a polynomial (thus of finite degree) and is nonzero, has no real roots.
Note that is not constant. We then find two complex roots: . We find that , and that . This means that . Thus, are roots of the polynomial, and so will be a factor of the polynomial. (Note: This requires the assumption that . Clearly, , because that would imply the existence of a real root.)
The polynomial is thus in the form of . Substituting into the given expression, we have
Thus either is 0 for any , or satisfies the same constraints as . Continuing, by infinite descent, for some .
Since for some , we have ; so .
Comment:
The answer is clearly correct, but the proof has a gap, i.e. there is no reason that . Since has no real roots, the degree must be even. Consider . Then since is non-zero, . Now the function applied repeatedly from some real starting value of x becomes arbitrarily large, and the limit of as approaches infinity is 1, so =1 for all x, or . Then for some polynomial , and . Now suppose h has degree m. It is clearly monic. Assume that the next highest non-zero coefficient in h is k. Then, subtracting from both sides of the equation yields a polynomial equality with degree on the left and degree on the right, a contradiction. So , and .
See also
2007 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.