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Difference between revisions of "Mock AIME II 2012 Problems/Problem 11"

(Created page with "==Problem== There exist real values of <math>a</math> and <math>b</math> such that <math>a+b=n</math>, <math>a^2+b^2=2n</math>, and <math>a^3+b^3=3n</math> for some value of <ma...")
 
 
(One intermediate revision by the same user not shown)
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==Solution==
 
==Solution==
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First, if <math>n=0</math>, then <math>a^2+b^2=0\implies a=b=0</math>. We now assume that <math>n\ne 0</math>.
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First, if <math>n=0</math>, then <math>a^2+b^2=0\implies a=b=0</math>. We now assume that <math>n\ne 0</math>.
 
Now, note that <math>2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab</math>.
 
Now, note that <math>2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab</math>.
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Also, we have <math>(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies 2n+2(2n-3)=n^2\implies n^2-6n+6=0</math>.
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Also, we have <math>(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies </math>
 +
<math>2n+2(2n-3)=n^2\implies n^2-6n+6=0</math>.
  
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Next, <math>3n^2=(a+b)(a^3+b^3)=a^4+b^4+ab(a^2+b^2)=a^4+b^4+(2n-3)(2n)\implies a^4+b^4=-n^2+6n</math>. But we know <math>n^2-6n+6=0</math>, so <math>-n^2+6n=6</math>.
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Next, <math>3n^2=(a+b)(a^3+b^3)=a^4+b^4+ab(a^2+b^2)=a^4+b^4+(2n-3)(2n)</math>
 +
<math>\implies a^4+b^4=-n^2+6n</math>. But we know <math>n^2-6n+6=0</math>, so <math>-n^2+6n=6</math>.
  
 
Since the only possible values of <math>a^4+b^4</math> are <math>0</math> and <math>6</math>, our final answer is <math>\boxed{006}</math>.
 
Since the only possible values of <math>a^4+b^4</math> are <math>0</math> and <math>6</math>, our final answer is <math>\boxed{006}</math>.
  
 
(It is easy to check that there exists <math>a, b, n</math> satisfying the equations.)
 
(It is easy to check that there exists <math>a, b, n</math> satisfying the equations.)

Latest revision as of 02:18, 5 April 2012

Problem

There exist real values of $a$ and $b$ such that $a+b=n$, $a^2+b^2=2n$, and $a^3+b^3=3n$ for some value of $n$. Let $S$ be the sum of all possible values of $a^4+b^4$. Find $S$.

Solution

First, if $n=0$, then $a^2+b^2=0\implies a=b=0$. We now assume that $n\ne 0$. Now, note that $2n^2=(a^2+b^2)(a+b)=a^3+b^3+ab(a+b)=3n+abn\implies 2n-3=ab$. Also, we have $(a+b)^2=n^2\implies a^2+b^2+2ab=n^2\implies$ $2n+2(2n-3)=n^2\implies n^2-6n+6=0$.

Next, $3n^2=(a+b)(a^3+b^3)=a^4+b^4+ab(a^2+b^2)=a^4+b^4+(2n-3)(2n)$ $\implies a^4+b^4=-n^2+6n$. But we know $n^2-6n+6=0$, so $-n^2+6n=6$.

Since the only possible values of $a^4+b^4$ are $0$ and $6$, our final answer is $\boxed{006}$.

(It is easy to check that there exists $a, b, n$ satisfying the equations.)

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