IMO 2007 Problem 5 (with the great IDMasterz)
by Binomial-theorem, Feb 12, 2013, 7:57 AM
Let
and
be positive integers. Show that if
divides
, then
.
Solution (with IDMasterz)





Solution (with IDMasterz)
First off, note that
is the same condition as
because
, therefore we get
. Note that
, and
. Therefore, the condition is the same as
.
Let
be a solution to
with WLOG
(because the equation is symmetric). Assume
is minimal, and I will prove this is absurd. Let
where
is another solution, therefore
. This is the same as
. Assume that
is negative or zero. However,
in this case, absurd. Therefore
is positive.
Now,
from Vieta's. Therefore,
.
. If
then this inequality is not true, and the following contradiction does not hold. However, when
, we get
which is clearly false, therefore
contradicting the minimality. Therefore,
is the only possible case, which is easy to check that it works. 







Let











Now,








