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Difference between revisions of "2015 USAMO Problems/Problem 5"

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Note:  This solution is definitely not what the folks at MAA intended, but it works!
 
Note:  This solution is definitely not what the folks at MAA intended, but it works!
  
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Look at the statement <math>a^4+b^4=e^5</math>.  This can be viewed as a special case of [[http://en.wikipedia.org/wiki/Beal%27s_conjecture|Beal's Conjecture]], stating that the equation <math>A^x+B^y=C^z</math> has no solutions over positive integers for <math>gcd(a, b, c) = 1</math> and <math>x, y, z > 2</math>. This special case was proven in 2009 by Michael Bennet, Jordan Ellenberg, and Nathan Ng, as <math>(x, y, z) = (2, 4, n)</math>.  This case <math>a^4+b^4=e^5</math> is obviously contained under that special case, so <math>a</math> and <math>b</math> must have a common factor greater than <math>1</math>.   
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Look at the statement <math>a^4+b^4=e^5</math>.  This can be viewed as a special case of Beal's Conjecture (http://en.wikipedia.org/wiki/Beal%27s_conjecture), stating that the equation <math>A^x+B^y=C^z</math> has no solutions over positive integers for <math>gcd(a, b, c) = 1</math> and <math>x, y, z > 2</math>. This special case was proven in 2009 by Michael Bennet, Jordan Ellenberg, and Nathan Ng, as <math>(x, y, z) = (2, 4, n)</math>.  This case <math>a^4+b^4=e^5</math> is obviously contained under that special case, so <math>a</math> and <math>b</math> must have a common factor greater than <math>1</math>.   
  
 
Call the greatest common factor of <math>a</math> and <math>b</math> <math>f</math>.  Then <math>a = f \cdot a_1</math> for some <math>a_1</math> and likewise <math>b = f \cdot b_1</math> for some <math>b_1</math>.  Then consider the quantity <math>ac+bd</math>.   
 
Call the greatest common factor of <math>a</math> and <math>b</math> <math>f</math>.  Then <math>a = f \cdot a_1</math> for some <math>a_1</math> and likewise <math>b = f \cdot b_1</math> for some <math>b_1</math>.  Then consider the quantity <math>ac+bd</math>.   

Revision as of 14:56, 13 May 2015

Problem

Let $a, b, c, d, e$ be distinct positive integers such that $a^4 + b^4 = c^4 + d^4 = e^5$. Show that $ac + bd$ is a composite number.

Solution

Note: This solution is definitely not what the folks at MAA intended, but it works!

Look at the statement $a^4+b^4=e^5$. This can be viewed as a special case of Beal's Conjecture (http://en.wikipedia.org/wiki/Beal%27s_conjecture), stating that the equation $A^x+B^y=C^z$ has no solutions over positive integers for $gcd(a, b, c) = 1$ and $x, y, z > 2$. This special case was proven in 2009 by Michael Bennet, Jordan Ellenberg, and Nathan Ng, as $(x, y, z) = (2, 4, n)$. This case $a^4+b^4=e^5$ is obviously contained under that special case, so $a$ and $b$ must have a common factor greater than $1$.

Call the greatest common factor of $a$ and $b$ $f$. Then $a = f \cdot a_1$ for some $a_1$ and likewise $b = f \cdot b_1$ for some $b_1$. Then consider the quantity $ac+bd$.

$ac+bd = f \cdot a_1 \cdot c + f \cdot b_1 \cdot d = f\cdot(a_1 \cdot c + b_1 \cdot d)$.

Because $b$ and $d$ are both positive, $(a_1 \cdot c + b_1 \cdot d) > 1$, and by definition $f > 1$, so $ac+bd$ is composite.

~BealsConjecture

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