Difference between revisions of "Mock AIME I 2012 Problems/Problem 12"
(Created page with "==Problem== Let <math>P(x)</math> be a polynomial of degree 10 satisfying <math>P(x^2) = P(x)P(x-1)</math>. Find the maximum possible sum of the coefficients of <math>P(x)</math>...") |
Viperstrike (talk | contribs) (→Solution) |
||
Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
Notice that if <math>a</math> is a root of <math>P</math>, then <math>a^2</math> must be a root of <math>P</math> and <math>(a + 1)^2</math> must be a root of <math>P</math>. But then continuing this, <math>a^{2^n}</math> and <math>(a + 1)^{2^n}</math> must be roots of <math>P</math> for all <math>n</math>. Since a polynomial has finitely many roots, <math>a</math> and <math>a + 1</math> must be roots of unity so that the above two sets contain finitely many elements. But there is a unique pair of roots of unity with real parts that differ by <math>1</math>, making <math>a = </math> <math>-1/2 \pm i\sqrt{3}/2</math>. Then the disjoint union of the two sets above is <math>\{-1/2 + i\sqrt{3}/2, 1/2 + i\sqrt{3}/2\}</math>, the minimal polynomial for which is <math>x^2 + x + 1</math>. Since any power of this base polynomial will work, <math>P(x) = (x^2 + x + 1)^5</math>, making the sum of coefficients <math>\boxed{243}</math>. | Notice that if <math>a</math> is a root of <math>P</math>, then <math>a^2</math> must be a root of <math>P</math> and <math>(a + 1)^2</math> must be a root of <math>P</math>. But then continuing this, <math>a^{2^n}</math> and <math>(a + 1)^{2^n}</math> must be roots of <math>P</math> for all <math>n</math>. Since a polynomial has finitely many roots, <math>a</math> and <math>a + 1</math> must be roots of unity so that the above two sets contain finitely many elements. But there is a unique pair of roots of unity with real parts that differ by <math>1</math>, making <math>a = </math> <math>-1/2 \pm i\sqrt{3}/2</math>. Then the disjoint union of the two sets above is <math>\{-1/2 + i\sqrt{3}/2, 1/2 + i\sqrt{3}/2\}</math>, the minimal polynomial for which is <math>x^2 + x + 1</math>. Since any power of this base polynomial will work, <math>P(x) = (x^2 + x + 1)^5</math>, making the sum of coefficients <math>\boxed{243}</math>. | ||
+ | |||
+ | |||
+ | -------------------------- | ||
+ | |||
+ | |||
+ | FALSE THE ABOVE IS FALSE. "UNIQUE PAIR OF ROOTS OF UNITY WITH REAL PARTS THAT DIFFER BY 1" IS FALSE. CONSIDER (<math>i,1</math>) for example.!!!!!! |
Revision as of 21:25, 15 March 2015
Problem
Let be a polynomial of degree 10 satisfying
. Find the maximum possible sum of the coefficients of
.
Solution
Notice that if is a root of
, then
must be a root of
and
must be a root of
. But then continuing this,
and
must be roots of
for all
. Since a polynomial has finitely many roots,
and
must be roots of unity so that the above two sets contain finitely many elements. But there is a unique pair of roots of unity with real parts that differ by
, making
. Then the disjoint union of the two sets above is
, the minimal polynomial for which is
. Since any power of this base polynomial will work,
, making the sum of coefficients
.
FALSE THE ABOVE IS FALSE. "UNIQUE PAIR OF ROOTS OF UNITY WITH REAL PARTS THAT DIFFER BY 1" IS FALSE. CONSIDER () for example.!!!!!!