2015 USAMO Problems/Problem 5

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 $ac+bd$ .