The article presents different ways of solving functional equations that occur in olimpiads and contests.
В статье представлены разные способы решения функциональных равенств, которые встречаются на олимпиадах и на конкурсах.
-
Find all functions
such that
is nonzero and
holds for all
[1].
Solution: Let be the assertion
.
From we get
:
and
, so

By subtracting the second equation from the first we get
(*). Now: Let
be the assertion
We have
, so
Again by subtracting the second equation from the first we get:
Now for
,
We know natural divisors of
are
Then
must be equal one of these
. So we consider eight cases:
1) Suppose .
In (*) from ,
is clear and
such
prime number is not exist.
2) Suppose . From (*),
.
is clear and in
we have
such
prime number is not exist.
3) Suppose . From (*),
In ,
Contradiction.
is satisfied.
is not satisfied.
For





4) Suppose . In (*),
For any prime
,
.
such
prime number is not exist.
5) . Such as can not be.
6)Suppose . In (*),
But for any prime
,
.
such
prime number is not exist
7) . Such as can not be.
So for any ,
Now from
. From this we get
By subtracting the second equation from the first we get:
And also we have

Now assume
. Then
and for
prime numbers,
From this
But for enough large
- prime number this is not true, because,
such
- prime number infinite. Contradiction. So we conclude,
such
is not exist.
For
,
-
Find all functions
such that
holds for all
[2].
Solution: be the assertion
.
:
Now consider this set
. Values of elements of
are bounded, lies between
and
. Let
the smallest. We have
.
. But because of
the smallest,
so for
. Now assume
.
:
Also we have
In this for
(
Then .
this number is positive and costanta, but
is unbounded. Contradiction. So we get
. From this equality we conclude the biggest element’s value and the smallest element’s value of
are equal.
All element’s value of
equal to
.
.
:
and
toq
For
,
from this,
So, But
unbounded, but
is positive
Contradiction. So, such
is not exist.
,
This is indeed a solution.
-
Find all functions
such that for all
[3].
Solution:


: (here
)
.
:
From this Now for
,
(**).
:
.
:
.
:
:
From (**).
:
For
. (
is clear).
for all
. This is indeed a solution
-
Find all functions
for all
[3].
Solution: be the assertion
.
:
injective and
:
. In this equality
:
.
:
Because of
injective:
or
If
, from
Contradiction
We have the following equalities:

:
:
(in (**)
:
). Because of
injective:
. In this equality
:
. From (***):
.
we can write (***):
:
(****)
:
.
.
-
Find all functions
for all
[1].
.
Solution: be the assertion
(*).
:
:
In this equality
. using this equality we can rewrite (*).

:
:
2
. In this equality
:
Now using this and (**) we get
. So
.
(****). Assume
:
(
:
).
from (***)
:
But from (****)
or
Contradiction.
Such
is not exist.
,
-
Find all functions
such that for
the following equalities hold:
,
.
Yechim: be the assertion
.
:
:
(*).
Let





:
:
Contradiction.
-case: suppose
. In (*)
:
So,
If
then from
we get
. So
If
, according to second condition,
.
-case:
.
:
.
:
:
Using this and (**), for
,
(***).Using this we can rewrite initial equality:
from (***)
For
:
So,
In (****) :
,
From (***), for
,
References:
- www.artofproblemsolving.com/community/c482986. IMO Shortlisted problems 2016.
- Mohammad Mahdi Taheri. “Functional equations in mathematical competitions: Problems and solutions” July 1, 2015.
- Ozgur Kircak. “Functional equations” April 8, 2011.