# 2015 USAMO Problems/Problem 5

### Problem

Let be distinct positive integers such that . Show that is a composite number.

### Solution 1

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

Look at the statement . This can be viewed as a special case of Beal's Conjecture (http://en.wikipedia.org/wiki/Beal%27s_conjecture), stating that the equation has no solutions over positive integers for and . This special case was proven in 2009 by Michael Bennet, Jordan Ellenberg, and Nathan Ng, as . This case is obviously contained under that special case, so and must have a common factor greater than .

Call the greatest common factor of and . Then for some and likewise for some . Then consider the quantity .

.

Because and are both positive, , and by definition , so is composite.

~BealsConjecture

### Solution 2

A more conventional approach, using proof by contradiction:

Without loss of generality, assume

Since , It is obvious that

We construct the equation (the right side is the factorization of the left) Which factors into: (by using and factoring)

If or equal zero, then equals or respectively, which is impossible because and because are distinct. Therefore the right side of the equation above is non-zero and the left side must be divisible by

If were prime,

We have

Which means that and neither nor can be multiples of meaning must be a multiple of and for some integer

But clearly this is impossible since .

Therefore, by contradiction, is composite.