2015 USAMO Problems/Problem 5
Let be distinct positive integers such that . Show that is a composite number.
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.
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,
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.