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 "2015 USAMO Problems/Problem 5"

(→‎Solution)
(→‎Solution)
Line 13: Line 13:
 
<math>ac+bd = f \cdot a_1 \cdot c + f \cdot b_1 \cdot d = f\cdot(a_1 \cdot c + b_1 \cdot d)</math>.
 
<math>ac+bd = f \cdot a_1 \cdot c + f \cdot b_1 \cdot d = f\cdot(a_1 \cdot c + b_1 \cdot d)</math>.
  
−
Because <math>b</math> and <math>d</math> are both positive, <math>(a_1 \cdot c + b_1 \cdot d) > 1</math>, and by definition <math>f > 1</math>, so <math>ac+bd</math> is composite.   
+
Because <math>c</math> and <math>d</math> are both positive, <math>(a_1 \cdot c + b_1 \cdot d) > 1</math>, and by definition <math>f > 1</math>, so <math>ac+bd</math> is composite.   
  
 
~BealsConjecture
 
~BealsConjecture

Revision as of 08:53, 29 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 $c$ 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_USAMO_Problems/Problem_5&oldid=70434"