Difference between revisions of "2006 IMO Problems/Problem 4"
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If <math>(x,y)</math> is a solution then obviously <math>x\geq 0</math> and <math>(x,-y)</math> is a solution too. For <math>x=0</math> we get the two solutions <math>(0,2)</math> and <math>(0,-2)</math>. | If <math>(x,y)</math> is a solution then obviously <math>x\geq 0</math> and <math>(x,-y)</math> is a solution too. For <math>x=0</math> we get the two solutions <math>(0,2)</math> and <math>(0,-2)</math>. | ||
Revision as of 13:59, 31 July 2019
Problem
Determine all pairs of integers such that
Solution
If is a solution then obviously
and
is a solution too. For
we get the two solutions
and
.
Now let be a solution with
; without loss of generality confine attention to
. The equation rewritten as
shows that the factors
and
are even, exactly one of them divisible by
. Hence
and one of these factors is divisible by
but not by
. So
Plugging this into the original equation we obtain
or, equivalently
Therefore
For
this yields
, i.e.
, which fails to satisfy
. For
equation
gives us
implying
. Hence
; on the other hand
cannot be
by
. Because
is odd, we obtain
, leading to
. From
we get
. These values indeed satisfy the given equation. Recall that then
is also good. Thus we have the complete list of solutions
:
,
,
,
.