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=SP-7LgQh0uY [Video contains solution to problem 3]
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
,
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
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 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 provide the solution of all infinite sequences of positive integers as:
,