Difference between revisions of "2015 USAMO Problems/Problem 5"
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<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> | + | 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 09:53, 29 May 2015
Problem
Let be distinct positive integers such that
. Show that
is a composite number.
Solution
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