2023 IMO Problems/Problem 3
Problem
For each integer , determine all infinite sequences of positive integers
for which there exists a polynomial
of the form
, where
are non-negative integers, such that
for every integer
.
Solution
https://www.youtube.com/watch?v=JhThDz0H7cI [Video contains solutions to all day 1 problems]
https://www.youtube.com/watch?v=CmJn5FKxpPY [Video contains another solution to problem 3]
Let and
be functions of positive integers
and
respectively.
Let , then
, and
Let
If we want the coefficients of to be positive, then
for all
which will give the following value for
:
Thus for every and
we need the following:
Solving for we get:
for all
and
because
needs to be greater than or equal to zero for all coefficients to be non-negative.
This means that needs to be increasing with
, or staying constant, and also with
because
.
In addition, since we need all coefficients to be integer, then all and
must also be integers. We also need
to not be dependent of
, so in the expression
, the
needs to cancel. This mean that the rate of change for
with respect to
needs to be constant. This can only be achieved with
be the equation of a line with slope being either zero or positive integer.
So, we set to be the equation of a line as
with
being the slope with a non-negative value and with
the intercept at
. We know that
so
which means that
and our function becomes
. Since
needs to be non-negative integer then
then
is increasing or constant, with
Then,
This gives:
with and coefficients of polynomial
Then,
Which provides the solution of all infinite sequences of positive integers as:
,
and
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2023 IMO (Problems) • Resources | ||
Preceded by Problem 6 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |