1996 IMO Problems/Problem 4
Contents
[hide]Problem
The positive integers and
are such that the numbers
and
are both squares of positive integers. What is the least possible value that can be taken on by the smaller of these two squares?
Solution
Let us first set up two equations given the information in the problem:
Let
Let
In general, when we are given alternating sums and differences like this, it is a good idea to square both equations and add them up to cancel out any cross terms. Doing this results in the equation .
Note that factorizes into
. Thus,
, and
. Now, we make a table for all modular congruences possible for m^4 for all nonnegative integers with a unique modulo 13 and 37. Note that due to the modular symmetry of even powers (
), we only need to consider up to
and
:
m ||
0 | 0 | 0 1 | 1 | 1 2 | 16 | 3 3 | 7 | 3 4 | 34 | 9 5 | 33 | 1 6 | 1 | 9 7 | 33 | 9 8 | 26 9 | 12 10| 10 11| 26 12| 16 13| 34 14| 10 15| 9 16| 9 17| 12 18| 7
Note that the purpose of this table is to find an two and
such that they add up to
for both
and
. Looking at
, this is clearly impossible unless both
and
are congruent to
. Thus,
. We do the same for
, and after a bit of calculation, we see that it is also impossible for
and
to be anything other than
. Thus,
and
are both divisible by
, or
. Logically, the first pair we test is
, and
. Indeed, this pair works with
and
. Since we know that
with
, we know that this is the smallest possible value to work. Thus, the minimum of
is
, or
.
~Stead (first IMO problem solution! :D)
Video Solution
https://www.youtube.com/watch?v=d3Olg8LekzA&t=5s
See Also
1996 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |