Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2007 AIME II Problems/Problem 14"

m (non-rigorous solution)
(Formalization! Yay!)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
: ''Note:The following solution(s) are non-rigorous.''
+
Let <math>r</math> be a root of <math>f(x)</math>. Then we have <math>f(r)f(2r^2)=f(2r^3+r)</math>; since <math>r</math> is a root, we have <math>f(r)=0</math>; therefore <math>2r^3+r</math> is also a root. Thus, if <math>r</math> is real and non-zero, <math>f(x)</math> has infinitely many roots. Since <math>f(x)</math> is a polynomial (thus of finite degree) and <math>f(0)</math> is nonzero, <math>f(x)</math> has no real roots.
  
−
Substitute the values <math>x = \pm i</math>. We find that <math>\displaystyle f(i)f(-2) = f(-i)</math>, and that <math>\displaystyle f(-i)f(-2) = f(i)</math>. This means that <math>f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2) - 1) = 0</math>. This suggests that <math>\pm i</math> are roots of the polynomial, and so <math>\displaystyle (x - i)(x + i) = x^2 + 1</math> will be a root of the polynomial.  
+
We then find two complex roots: <math>r = \pm i</math>. We find that <math>f(i)f(-2) = f(-i)</math>, and that <math>f(-i)f(-2) = f(i)</math>. This means that <math>f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2) - 1) = 0</math>. Thus <math>\pm i</math> are roots of the polynomial, and so <math>(x - i)(x + i) = x^2 + 1</math> will be a factor of the polynomial.  
  
−
The polynomial is likely in the form of <math>f(x) = (x^2 + 1)g(x)</math>; <math>g(x)</math> appears to satisfy the same relation as <math>f(x)</math>, so it also probably has the same roots. Thus, <math>f(x) = (x^2 + 1)^nh(x)</math> is the solution. Guessing values for <math>h(x)</math>, try <math>h(x) = 1</math>. Checking a couple of values shows that <math>f(x) = (x^2 + 1)^2</math> works, and so the solution is <math>f(5) = 676</math>.
+
The polynomial is thus in the form of <math>f(x) = (x^2 + 1)g(x)</math>. Substituting into the given expression, we have
 +
 
 +
<cmath>(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)</cmath>
 +
<cmath>(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)</cmath>
 +
 
 +
Thus either <math>4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)</math> is 0 for any <math>x</math>, or <math>g(x)</math> satisfies the same constraints as <math>f(x)</math>. Continuing, by infinite descent, <math>f(x) = (x^2 + 1)^n</math> for some <math>n</math>.
 +
 
 +
Since <math>f(2)+f(3)=125=5^n+10^n</math> for some <math>n</math>, we have <math>n=2</math>; so <math>f(5) = 676</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:46, 12 March 2009

Problem

Let $f(x)$ be a polynomial with real coefficients such that $\displaystyle f(0) = 1,$ $\displaystyle f(2)+f(3)=125,$ and for all $x$, $\displaystyle f(x)f(2x^{2})=f(2x^{3}+x).$ Find $\displaystyle f(5).$

Solution

Let $r$ be a root of $f(x)$. Then we have $f(r)f(2r^2)=f(2r^3+r)$; since $r$ is a root, we have $f(r)=0$; therefore $2r^3+r$ is also a root. Thus, if $r$ is real and non-zero, $f(x)$ has infinitely many roots. Since $f(x)$ is a polynomial (thus of finite degree) and $f(0)$ is nonzero, $f(x)$ has no real roots.

We then find two complex roots: $r = \pm i$. We find that $f(i)f(-2) = f(-i)$, and that $f(-i)f(-2) = f(i)$. This means that $f(i)f(-i)f(-2)^2 = f(i)f(-i) \Longrightarrow f(i)f(-i)(f(-2) - 1) = 0$. Thus $\pm i$ are roots of the polynomial, and so $(x - i)(x + i) = x^2 + 1$ will be a factor of the polynomial.

The polynomial is thus in the form of $f(x) = (x^2 + 1)g(x)$. Substituting into the given expression, we have

\[(x^2+1)g(x)(4x^4+1)g(2x^2)=((2x^3+x)^2+1)g(2x^3+x)\] \[(4x^6+4x^4+x^2+1)g(x)g(2x^2)=(4x^6+4x^4+x^2+1)g(2x^3+x)\]

Thus either $4x^6+4x^4+x^2+1=(4x^4+1)(x^2+1)$ is 0 for any $x$, or $g(x)$ satisfies the same constraints as $f(x)$. Continuing, by infinite descent, $f(x) = (x^2 + 1)^n$ for some $n$.

Since $f(2)+f(3)=125=5^n+10^n$ for some $n$, we have $n=2$; so $f(5) = 676$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2007_AIME_II_Problems/Problem_14&oldid=30679"