Difference between revisions of "2023 IMO Problems/Problem 3"
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This means that <math>f(n)</math> needs to be increasing with n or staying constant with <math>f(1)=0</math> because <math>a_{1}=a_{1}+f(1)</math> In addition, since we need all coefficients to be integer then all <math>f(n)</math> and <math>g(j)</math> must also be integers | This means that <math>f(n)</math> needs to be increasing with n or staying constant with <math>f(1)=0</math> because <math>a_{1}=a_{1}+f(1)</math> In addition, since we need all coefficients to be integer then all <math>f(n)</math> and <math>g(j)</math> must also be integers | ||
| − | So, if we set <math>f(n)=(n-1)d</math> with <math>d\ | + | So, if we set <math>f(n)=(n-1)d</math> with <math>d\geq 0 \mid d \in \mathbb{Z}</math> then <math>f(n)</math> is increasing or constant, with <math>f(1)=0</math> |
Then, <math>g(j)=f(n+j)-f(n)=(n+j-1)d-(n-1)d=jd</math> | Then, <math>g(j)=f(n+j)-f(n)=(n+j-1)d-(n-1)d=jd</math> | ||
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This gives: | This gives: | ||
<math>P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!d^{k}</math> | <math>P(a_{n})=a_{n}^{k}+C_{k-1}a_{n}^{k-1}+...+C_{1}a_{n}+k!d^{k}</math> | ||
| + | |||
| + | with <math>C_{0}=k!d^{k}</math> and coefficients of polynomial <math>\geq 0</math> | ||
| + | |||
| + | Then, <math>a_{n}=a_{1}+f(n)=a_{1}+(n-1)d</math>, <math>\forall d\geq 0 \mid d \in \mathbb{Z}</math> | ||
Revision as of 12:20, 3 October 2023
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 n and i 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 n or staying constant with
because
In addition, since we need all coefficients to be integer then all
and
must also be integers
So, if we set
with
then
is increasing or constant, with
Then,
This gives:
with
and coefficients of polynomial
Then,
,