Difference between revisions of "2015 USAMO Problems/Problem 5"
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===Solution 2=== | ===Solution 2=== |
Revision as of 11:56, 30 December 2024
Contents
[hide]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
-->The result mentioned by author of solution is valid only for n>=6, which makes solution invalid
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.