1984 IMO Problems/Problem 6
Contents
Problem
Let be odd integers such that
and
. Prove that if
and
for some integers
and
, then
.
Solution 1
Let . As
, we infer that
; in particular,
.
Now, , hence
. It is easy to see that for
, if
, then
. If
,
, so
, which is in contradiction with the fact that
. Thereby,
.
Write . If
, contradiction; so
and
, or equivalently
(
otherwise
or
, contradiction )
Substituting back,
As and
is odd, we get that
. Furthermore,
, hence
. Now
which together with the fact that
, we get that
, so the family of the solutions
is described by the set
This solution was posted and copyrighted by TheFunkyRabbit. The original thread for this problem can be found here: [1]
Solution 2
Note that, if , then using
, and
, we get
, which is clearly impossible. Thus,
must hold.
Next, observe that, and
implies
. Keeping this in mind, we now observe that,
, implies
. Thus,
. Now, note also that, if
, then
, and thus, the largest power of
dividing either
or
is exactly
. Clearly,
, and thus,
. Moreover, if
, then we again have a contradiction, as
. Thus,
. This yields,
, which brings us,
. Now, using
, we immediately obtain
. This immediately establishes (since
is odd), that
.
This solution was posted and copyrighted by grupyorum. The original thread for this problem can be found here: [2]
See Also
1979 IMO (Problems) • Resources | ||
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