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r(s) and The IMO Compendium Group Functional Equations Marko Radovanovi´ c'[email protected] Co

- 2007 The Author(s) and The IMO Compendium Group

Problems for Independent Study

Basic Methods For Solving Functional Equations • Substituting the values for variables

The most common first attempt is with some constants (eg

0 or 1),

after that (if possible) some expressions which will make some part of the equation to become constant

- • Mathematical induction

especially with easier problems

• Investigating for injectivity or surjectivity of functions involved in the equaiton

In many of the problems these facts are not difficult to establish but can be of great importance

• Finding the fixed points or zeroes of functions

The number of problems using this method is considerably smaller than the number of problems using some of the previous three methods

This method is mostly encountered in more difficult problems

• Using the Cauchy’s equation and equation of its type

• Investigating the monotonicity and continuity of a function

Continuity is usually given as additional condition and as the monotonicity it usually serves for reducing the problem to Cauchy’s equation

the problem is on the other side of difficulty line

• Assuming that the function at some point is greater or smaller then the value of the function for which we want to prove that is the solution

Most often it is used as continuation of the method of mathematical induction and in the problems in which the range is bounded from either side

• Making recurrent relations

This method is usually used with the equations in which the range is bounded and in the case when we are able to find a relashionship between f ( f (n)),

- imomath

• Analyzing the set of values for which the function is equal to the assumed solution

The goal is to prove that the described set is precisely the domain of the function

- • Substituting the function

• Expressing functions as sums of odd and even

Namely each function can be represented as a sum of one even and one odd function and this can be very handy in treating ”linear” functional equations involving many functions

• Treating numbers in a system with basis different than 10

Of course,

this can be used only if the domain is N

• For the end let us emphasize that it is very important to guess the solution at the beginning

This can help a lot in finding the appropriate substitutions

- at the end of the solution,

DON’T FORGET to verify that your solution satisfies the given condition

If its domain is Q,

it is wellknown that the solution is given by f (x) = x f (1)

With a relatively easy counter-example we can show that the solution to the Cauchy equation in this case doesn’t have to be f (x) = x f (1)

However there are many additional assumptions that forces the general solution to be of the described form

Namely if a function f satisfies any of the conditions: • monotonicity on some interval of the real line

- • continuity

• boundedness on some interval

• positivity on the ray x ≥ 0

then the general solution to the Cauchy equation f : R → S has to be f (x) = x f (1)

The following equations can be easily reduced to the Cauchy equation

• All continuous functions f : R → (0,

+∞) satisfying f (x + y) = f (x) f (y) are of the form f (x) = ax

• All continuous functions f : (0,

+∞) → R satisfying f (xy) = f (x) + f (y) are of the form f (x) = loga x

Now the function g(x) = f (ax ) is continuous and satisfies the Cauchy equation

• All continuous functions f : (0,

- +∞) → (0,

+∞) satisfying f (xy) = f (x) f (y) are f (x) = xt ,

where t = loga b and f (a) = b

The following examples should illustrate the previously outlined methods

Problem 1

Marko Radovanovi´c: Functional Equations

Solution

and since f (1) = 2 we have f (n) = n + 1 for every natural number n

and setting x = −1 and y = n gives f (−n) = − f (n − 1) + 1 = −n + 1

(1) f (1) = (n + 1) f n n 1 1 1 Furthermore for x = 1 and y = m + we get f m + 1 + = f m+ + 1,

hence by the mathen n n 1 1 = m+ f

- f n n m m 1 = + 1,

f (r) = for every natural number n

Furthermore for x = m and y = we get f n n n r + 1,

for every positive rational number r

Setting x = −1 and y = r we get f (−r) = − f (r − 1)+ 1 = −r + 1 as well hence f (x) = x + 1,

- for each x ∈ Q

- for all x,
- y ∈ Q,

f is the solution to our equation

- △ Problem 2

(Belarus 1997) Find all functions g : R → R such that for arbitrary real numbers x and y: g(x + y) + g(x)g(y) = g(xy) + g(x) + g(y)

Solution

Notice that g(x) = 0 and g(x) = 2 are obviously solutions to the given equation

Using mathematical induction it is not difficult to prove that if g is not equal to one of these two functions then g(x) = x for all x ∈ Q

where r is rational and x real number

Particularly from the second equation for r = −1 we get g(−x) = −g(x),

hence setting y = −x in the initial equation gives g(x)2 = g(x2 )

Now we use the standard method of extending to R

Choose r ∈ Q such that g(x) < r < x

Then r > g(x) = g(x − r) + r ≥ r,

which is clearly a contradiction

Similarly from g(x) > x we get another contradiction

- △ Problem 3

The function f : R → R satisfies x + f (x) = f ( f (x)) for every x ∈ R

Solution

The domain of this function is R,

so there isn’t much hope that this can be solved using mathematical induction

because of injectivity we must have f (0) = 0,

- implying f ( f (0)) = 0

injectivity would imply f (x) = f (0) and x = 0

- △ Problem 4

- (b) f (1) = 2,
- f (2) = 4

- imomath

- and m = n,

n = 1 afterwards we get f ( f (1) + f (n)) = f ( f (1)) + f (n),

f ( f (n) + f (1)) = f ( f (n)) + f (1)

Let us emphasize that this is one standard idea if the expression on one side is symmetric with respect to the variables while the expression on the other side is not

From here we conclude that f (n) = m implies f (m) = m + 2 and now the induction gives f (m + 2k) = m + 2k + 2,

- for every k ≥ 0

Specially if f (1) = 2 then f (2n) = 2n + 2 for all positive integers n

The injectivity of f gives that at odd numbers (except 1) the function has to take odd values

We have f (2p + 2s + 1) = 2p + 2s + 3 for s'≥ 0

- 2p − 1 are mapped into 1,

then for m = n = t 4 = f (2) = f ( f (t) + f (t)) = f ( f (t)) + f (t) = 3,

- which is a contradiction

If for some t such that f (t) = 3 then f (3 + 2k) = 5 + 2k,

which is a contradiction to the existence of such t

It follows that the numbers 3,

- 2p − 1 are mapped into 5,

- for n ≥ 2

- △ Problem 5
- (BMO 1997,
- 2000) Solve the functional equation f (x f (x) + f (y)) = y + f (x)2 ,
- y ∈ R

Solution

In probelms of this type it is usually easy to prove that the functions are injective or surjective,

if the functions are injective/surjective

In this case for x = 0 we get f ( f (y)) = y + f (0)2

Since the function on the right-hand side is surjective the same must hold for the function on the left-hand side

- f (0) = 0

hence f (x)2 = x2 for every real number x

and after taking squares (1 + y)2 = f (1 + f (y))2 = (1 + f (y))2 = 1 + 2 f (y) + f (y)2 = 1 + 2 f (y) + y2

and after taking squares (1 + y)2 = f (−1 + f (y))2 = (−1 + f (y))2 = 1 − 2 f (y) + f (y)2 = 1 − 2 f (y) + y2

Now we conclude f (y) = −y for every real number y

It is easy to verify that f (x) = x and f (x) = −x are indeed the solutions

- △ Problem 6
- (IMO 1979,

shortlist) Given a function f : R → R,

if for every two real numbers x and y the equality f (xy + x + y) = f (xy) + f (x) + f (y) holds,

prove that f (x + y) = f (x) + f (y) for every two real numbers x and y

This is a clasical example of the equation that solution is based on a careful choice of values that are plugged in a functional equation

Plugging in y = 1 we get f (2x + 1) = 2 f (x) + f (1) and hence f (2(u + v+ uv)+ 1) = 2 f (u + v+ uv)+ f (1) = 2 f (uv)+ 2 f (u)+ 2 f (v)+ f (1) for all real u and v

On the other hand,

plugging in x = u and y = 2v+1 we get f (2(u+v+uv)+1) = f (u+(2v+1)+u(2v+ 1)) = f (u) + 2 f (v) + f (1) + f (2uv + u)

f (2uv + u) = 2 f (uv) + f (u)

Plugging in v = −1/2 we get 0 = 2 f (−u/2)+ f (u) = −2 f (u/2)+ f (u)

f (u) = 2 f (u/2) and consequently f (2x) = 2 f (x) for all reals

Now (1) reduces to f (2uv + u) = f (2uv) + f (u)

Plugging in u = y and x = 2uv,

we obtain f (x) + f (y) = f (x + y) for all nonzero reals x and y

it trivially holds that f (x + y) = f (x) + f (y) when one of x and y is 0

Marko Radovanovi´c: Functional Equations

Problem 7

Does there exist a function f : R → R such that f ( f (x)) = x2 − 2 for every real number x

- ? Solution

Notice that the function g of the right-hand side has exactly 2 fixed points and that the function g ◦ g has exactly 4 fixed points

Assume the contrary

- b be the fixed points of g,
- d the fixed points of g ◦ g

Assume that g(c) = y

hence g(g(y)) = g(c) = y and y has to be on of the fixed points of g ◦ g

If y = a then from a = g(a) = g(y) = c'we get a contradiction

- and since y 6= c'we get y = d

Thus g(c) = d'and g(d) = c

- hence f (x0 ) ∈ {a,

Similarly if x1 ∈ {a,

- d} we get f (x1 ) ∈ {a,

and now we will prove that this is not possible

However we then have f (d) = f ( f (c)) = g(c) = d,

- which is a contradiction,

This proves that the required f doesn’t exist

- △ Problem 8

Find all functions f : R+ → R+ such that f (x) f (y f (x)) = f (x+ y) for every two positive real numbers x,

Obviously f (x) ≡ 1 is one solution to the problem

For every x such that ≥ 0 we can find such y and f (x) − 1 from the given condition we get f (x) = 1

However this is a contradiction since we got that f (x) > 1 implies f (x) = 1

Assume that f (x) < 1 for some x

- for each x

- a) (a > 0)

Substituting x = y = 3 contradiction

Again everything will revolve around the idea of getting rid of f (y f (x))

therefore f (x) f (y f (x)) = f (x + y) = f (y f (x) + x + y − y f (x)) = f (y f (x)) f f y f (x) (x + y − y f (x)) ,

- i

f (x) = f f y f (x) (x + y − y f (x))

The injectivity of f implies that x = f y f (x) (x + y − y f (x))

If we plug f (x) = a we get

- f (y) =

1 + αz

- 1 − f (a) ,

and according to our assumption α > 0

- a f (a) 1 ,
- for α ∈ R+ ,

and f (x) ≡ 1 satisfy the equation

△ It is easy to verify that f (x) = 1 + αx Problem 9

- (IMO 2000,

shortlist) Find all pairs of functions f : R → R and g : R → R such that for every two real numbers x,

y the following relation holds: where α =

f (x + g(y)) = x f (y) − y f (x) + g(x)

Solution

Setting y = α in the given equation yields g(x) = (α + 1) f (x) − x f (α )

so setting y = α + 1 we get f (x + n) = mx,

where n = g(α + 1) and m = f (α + 1) − f (α )

and consequently g is also linear

If we now substitute f (x) = ax + b and g(x) = cx + d'in the given equation and compare the coefficients,

- we easily find that f (x) =
- cx − c2 1+c
- and g(x) = cx − c2,
- c ∈ R \ {−1}

- imomath

- so we can take α = −g(0)

By replacing x by g(x) in the given equation we obtain f (g(x) + g(y)) = g(x) f (y) − y f (g(x)) + g(g(x)) and,

- analogously,

f (g(x) + g(y)) = g(y) f (x) − x f (g(y)) + g(g(y))

- where a = g(0)

g is injective and f is surjective,

so there exists c'∈ R such that f (c) = 0

Now the above two relations yield g(x) f (y) − ay + g(g(x)) = g(y) f (x) − ax + g(g(y))

(1) Plugging y = c'in (1) we get g(g(x)) = g(c) f (x) − ax + g(g(c)) + ac = k f (x) − ax + d

Now (1) becomes g(x) f (y) + k f (x) = g(y) f (x) + k f (y)

- whence a−k g(x) = f (x) + k

b Note that g(0) = a 6= k = g(c),

- since g is injective

- so it takes the value 0
- △ Problem 10
- (IMO 1992,

shortlist) Find all functions f : R+ → R+ which satisfy f ( f (x)) + a f (x) = b(a + b)x

where x0 ≥ 0 is a fixed real number

It follows from the given equation in f that xn+2 = −axn+1 + b(a + b)xn

- where λ1 ,

In order to have xn ≥ 0 for all n we must have λ2 = 0

we conclude that f (x) = bx is the only possible solution of the functional equation

It is easily verified that this is indeed a solution

- △ Problem 11

(Vietnam 2003) Let F be the set of all functions f : R+ → R+ which satisfy the inequality f (3x) ≥ f ( f (2x)) + x,

for every positive real number x

- x 1 Solution

- hence α ≤

The idea is the following: Denote = α1 and form a sequence {αn } for which 3 3 1 1 1 f (x) ≥ αn x and which will (hopefully) tend to

- and hence α =
- 2 2 2 Let us constract a recurrent relation for αk

Assume that f (x) ≥ αk x,

- for every x ∈ R+

From the given inequality we have f (3x) ≥ f ( f (2x)) + x ≥ αk f (2x) + x ≥ αk · αk · 2x + x = αk+1 · 3x

- 1 2αn2 + 1

This is a standard problem

It 3 2 1 is easy to prove that the sequence αk is increasing and bounded above by

Hence it converges and 2 2α 2 + 1 1 ,

α = (since α < 1)

△ its limit α satisfies α = 3 2

h : R → R that satisfy f (x + y) + g(x − y) = 2h(x) + 2h(y)

- where h(0) = b

Now the original equation can be written as 2 x − y x + y +h + h(x − y) + b = h(x) + h(y)

(2) 2 h 2 2 Let H(x) = h(x)− b

These ”longer” linear expressions can be easily handled if we express functions in form of the sum of an even and odd function,

H(x) = He (x) + Ho (x)

- y) and (x,

−y) we can add them together and get: x + y x − y 2 He + He + He (x − y) = He (x) + He (y)

(3) 2 2 If we set −y in this expression and add to (3) we get (using He (y) = He (−y)) He (x + y) − He(x − y) = 2He (x) + 2He (y)

The last equation is not very difficult

- for every rational number r

Similar method gives the simple relation for Ho Ho (x + y) + Ho(x − y) = 2Ho (x)

- g(x) = α x2 + a

It is easy to verify that these functions satisfy the given conditions

Find all functions f : Q → Q for which f (xy) = f (x) f (y) − f (x + y) + 1

Solution

- f (0) = 1

Furthermore,

setting x = 1 and y = −1 gives f (−1) = f (1) f (−1),

hence f (−1) = 0 or f (1) = 1

We will separate this into two cases: 1◦ Let f (−1) = 0

In this innocent-looking problems that are resistent to usual ideas it is sometimes successful to increase the number of variables,

to set yz instead of y: f (xyz) = f (x) f (yz) − f (x + yz) + 1 = f (x)( f (y) f (z) − f (y + z) + 1) − f (x + yz) + 1

Although it seems that the situation is worse and running out of control,

- that is not the case

Namely the expression on the left-hand side is symmetric,

while the one on the right-hand side is not

Writing the same expression for x and equating gives f (x) f (y + z) − f (x) + f (x + yz) = f (z) f (x + y) − f (z) + f (xy + z)

Setting z = −1 (we couldn’t do that at the beginning,

since z = 1 was fixed) we get f (x) f (z − 1) − f (x) + f (x − y) = f (xy − 1),

and setting x = 1 in this equality gives f (y − 1)(1 − f (1)) = f (1 − y) − f (1)

- f (1) = 0 or f (1) = 2

- imomath
- 1◦ If f (1) = 0,

then from (5) plugging y + 1 instead of y we get f (y) = f (−y)

- hence f (x+ y) = f (x − y),

for every two rational numbers x and y

- for all x ∈ Q

In this case we don’t have a solution

- 2◦ If f (1) = 2,

setting y + 1 instead of y in (5) gives 1 − f (y) = f (−y) − 1

It is clear that we should do the substitution g(x) = 1 − f (x) because the previous equality gives g(−x) = −g(x),

- g is odd

Furthermore substituting g into the original equality gives g(xy) = g(x) + g(y) − g(x)g(y) − g(x + y)

and adding with (6) yields g(x+ y)+ g(x− y) = 2g(x)

and easy verification shows that f (x) = 1 + x satisfies the conditions of the problem

- 2◦ Let f (1) = 1

hence for y = −1 we get f (1 − x) = 1,

- for every rational x

This means that f (x) ≡ 1 and this function satisfies the given equation

hence we conclude that for all rational numbers q f (q) = q + 1,

- or f (q) ≡ 1

it can be easily shown that f (x) ≡ 1

From the above we have that g(x) + g(y) = g(x + y),

hence it is enough to prove monotonicity

Substitute x = y in (6) and use g(2x) = 2g(x) to get g(x2 ) = −g(x)2

Hence if y > x,

y = x + r2 we have g(y) = g(x) + g(r2) ≤ g(x),

and the function is decreasing

This means that f (x) = 1 + α x and after some calculation we get f (x) = 1 + x

- △ Problem 14
- (IMO 2003,

shortlist) Let R+ denote the set of positive real numbers

Find all functions f : R+ → R+ that satisfy the following conditions: √ √ √ (i) f (xyz) + f (x) + f (y) + f (z) = f ( xy) f ( yz) f ( zx) (ii) f (x) < f (y) for all 1 ≤ x < y

First notice that the solution of this functional equation is not one of the common solutions 1 that we are used to work with

Namely one of the solutions is f (x) = x + which tells us that this x equality is unlikely to be shown reducing to the Cauchy equation

setting x = y = z = 1 we get f (1) = 2 (since f (1) > 0)

and proving this equality will be our next step

Putting x = ts,

- y = ts ,

z = ts in (i) gives f (t) f (s) = f (ts) + f (t/s)

for s'= 1 the last equality yields f (t) = f (1/t)

hence f (t) ≥ f (1) = 2 for each t

Now it follows by induction from (7) n n q that g(t ) = g(t) for every integer n,

and therefore g(t ) = g(t)q for every rational q

- if t > 1 is fixed,

we have f (t q ) = aq + a−q ,

- where a = g(t)

But since the set of aq (q ∈ Q) is dense in R+ and f is monotone on (0,

1] and [1,

it follows that f (t r ) = ar + a−r for every real r

- if k is such that t k = a,
- we have f (x) = xk + x−k
- for every x ∈ R

Marko Radovanovi´c: Functional Equations

Find all functions f : [1,

- ∞) → [1,

∞) that satisfy: (i) f (x) ≤ 2(1 + x) for every x ∈ [1,

(ii) x f (x + 1) = f (x)2 − 1 for every x ∈ [1,

Solution

Let us prove that this is the only solution

Using the given conditions we get f (x)2 = x f (x + 1) + 1 ≤ x(2(x + 1)) + 1 < 2(1 + x)2,

- f (x) ≤ 2(1 + x)

With this we have found the upper bound for f (x)

Since our goal is to prove f (x) = x + 1 we will use the same method for lowering the upper bound

- f (x) < 21/2 (1 + x),
- for every k

However this is shown in the same way as the previous two inequalities

hence for fixed x we can’t have f (x) > x + 1

It remains to show that f (x) ≥ x + 1,

- for x ≥ 1

√ f (x)2 − 1 = f (x + 1) ≥ 1,

- f (x) ≥ x + 1 > x1/2

From the fact that the range is [1,

+∞) we get x √ We further have f (x)2 = 1 + x f (x + 1) > 1 + x x + 2 > x3/2 and similarly by induction k

- f (x) > x1−1/2

- el f (x) ≥ x + 1/2

and 2 passing to the limit we get the required inequality f (x) ≥ x + 1

- △ Problem 16
- (IMO 1999,

probelm 6) Find all functions f : R → R such that f (x − f (y)) = f ( f (y)) + x f (y) + f (x) − 1

A = f (R)

- for some y

From the given equality we have f (0) = f (x) + x2 + f (x) − 1,

- f (x) =
- c + 1 x2 − ,
- where f (0) = c

Now it is clear that we have to analyze set A further

Setting x = y = 0 in the original equation we get f (−c) = f (c) + c'− 1,

- hence c'6= 0

plugging y = 0 in the original equation we get f (x − c) − f (x) = cx + f (c) − 1

we get { f (x − c) − f (x) | x ∈ R} = R,

A − A = R

Hence for every real number x there are real numbers y1 ,

- y2 ∈ A such that x = y1 − y2

f (y1 − y2) = f (y1 − f (z)) = f ( f (z)) + y1 f (z) + f (y1 ) − 1

f (y1 ) + f (y2 ) + y1 y2 − 1 = c'−

- x2 From the original equation we easily get c'= 1

Olympiad Training Materials,

- imomath

let f : R → R be a continuous function satisfying f (0) = 0,

- f (1) = 1,
- and f (n) (x) = x,
- for every x ∈ [0,

- hence f is injective

Let us prove that formally

that for some two real numbers x1 < x2 we have f (x1 ) ≥ f (x2 )

- x1 ] implies that there is some c'such that f (c) = f (x2 ),

which contradicts the injectivity of f

Now if x < f (x),

- we get f (x) < f ( f (x)) etc
- x < f (n)(x) = x

Similarly we get a contradiction if we assume that x > f (x)

Hence for each x ∈ [0,

- 1] we must have f (x) = x
- △ Problem 18

- +∞) → (0,

+∞) that satisfy f ( f (x) + y) = x f (1 + xy) for all x,

- y ∈ (0,
- 1 is one solution to the functional equation

Let us prove that the function x is non-increasing

We will y f (y) − x f (x) consider the expression of the form z = since it is positive and bigger then f (y)

- z − f (y)) instead of (x,
- y) in the original equation,

then we plug z − f (x) instead of y,

- we get x = y,
- which is a contradiction

Hence the function is non-decreasing

Substituting x = 1 we get f ( f (1)+ y) = f (1 + y),

hence f (u + | f (1) − 1|) = f (u) for u > 1

Therefore the function is periodic on the interval (1,

and since it is monotone it is constant

However we then conclude that the left-hand side of the original equation constant and the right-hand side is not

Indeed for y = 1 − the given equality gives f f (x) − x x x x 1 1 1 we have f f (x) − + 1 ≤ f (1) = 1 and x f (x) > 1

If f (x) < we have f f (x) − + 1 ≥ x x x 1 1 f (1) = 1,

- and x f (x) < 1

Hence f (x) =

If x < 1,

plugging y = we get x x 1 x f f (x) + = x f (2) = ,

x 2 1 2 1 1 1 and since ≥ 1,

- we get f (x) + = ,
- f (x) = in this case,

- △ Solution

Problem 19

(Bulgaria 1998) Prove that there is no function f : R+ → R+ such that f (x)2 ≥ f (x + y)( f (x) + y) for every two positive real numbers x and y

The common idea for the problems of this type is to prove that f (y) < 0 for some y > 0 which will lead us to the obvious contradiction

for every x because the simple addition gives f (x) − f (x + m) ≥ mc

- for every x

From the given inequality we get f (x) − f (x + y) ≥ f (x + y)y and the function is obviously decreasing

f (x) − f (x + y) ≥ f (x) + y Let n be a natural number such that f (x + 1)n ≥ 1 (such number clearly exists)

Notice that for 0 ≤ k ≤ n − 1 the following inequality holds f x + nk 1n k 1 k + 1 f x+ ≥ ,

− f x+ ≥ n n 2n f x+ k + 1 n

- 1 and adding similar realitions for all described k yields f (x) − f (x + 1) ≥ which is a contradiction
- 2 △ Problem 20

- f (2) = 1,
- f (3n) = 3 f (n),
- f (3n + 1) = 3 f (n) + 2,
- f (3n + 2) = 3 f (n) + 1

Let us calculate f (n) for n ≤ 8 in an attempt to guess the solution

- f ((1)3 ) = (2)3 ,
- f ((2)3 ) = (1)3 ,
- f ((10)3 ) = 6 = (20)3 ,
- f ((11)3 ) = 8 = (22)3 ,
- f ((12)3 ) = 7 = (21)3 ,
- f ((20)3 ) = 3 = (10)3 ,
- f ((21)3 ) = 5 = (12)3 ,
- f ((22)3 ) = 4 = (11)3

- and conversely

It is clear that f (n) = 2n if and only if in the system with base 3 n doesn’t contain any digit 1 (because this would imply f (n) < 2n)

Now it is easy to count the number of such n’s

- △ 2004 if f : Q → [0,
- +∞) is the Problem 21
- (BMO 2003,

shortlist) Find all possible values for f 2003 function satisfying the conditions: (i) f (xy) = f (x) f (y) for all x,

- y ∈ Q

(ii) f (x) ≤ 1 ⇒ f (x + 1) ≤ 1 for all x ∈ Q

- 2003 (iii) f = 2
- 2002 Solution

- for every rational x

Now (i) implies that for x = y = 1 we get f (1) = 0 and similarly for x =y = −1 we get f (−1) = 1

Byinduction y y f (y) = f (x) we have that f ≤ 1,

and f (x) ≤ 1 for every integer x

For f (x) ≤ f (y) from f x x y according to (ii) f + 1 ≤ 1

x hence f (x + y) ≤ max{ f (x),

- f (y)},
- for every x,
- y ∈ Q

there are integers a and b such that au + bv = 1

What is all of this good for

- ? We got that f (1) = 1,

and we know that f (x) ≤ 1 for all x ∈ Z and since 1 is the maximum of the function on Z and since we have the previous inequality our goal is to show that the value of the function is 1 for a bigger class of integers

If for every prime p we have f (p) = 1 then f (x) = 1 for every integer implying f (x) ≡ 1 which contradicts (iii)

Assume therefore that f (p) 6= 1 for some p ∈ P

There are a and b such that ap + bq = 1 implying f (1) = f (ap + bq) ≤ max{ f (ap),

- f (bq)}

Now we must have f (bq) = 1 implying that f (q) = 1 for every other prime number q

From (iii) we have 2003 f (2003) = = 2,

f 2002 f (2) f (7) f (11) f (13) hence only one of the numbers f (2),

- f (11),
- f (13) is equal to 1/2

- 2003 f (2003)

- imomath

2003 1 If f (2) = 1/2 then f = ,

- otherwise it is 1
- 2002 4 It remains to construct one function for each of the given values

and 1 for all other prime numbers

in the second case it is a the multiplicative function that takes the value 1/2 at,

- for example,
- 7 and takes 1 at all other prime numbers

For these functions we only need to verify the condition (ii),

but that is also very easy to verify

- △ Problem 22

Let I = [0,

G = I × I and k ∈ N

Find all f : G → I such that for all x,

z ∈ I the following statements hold: (i) f ( f (x,

- z) = f (x,
- (ii) f (x,

1) = x,

- y) = f (y,
- (ii) f (zx,
- zy) = zk f (x,
- y) for every x,
- z ∈ I,

where k is a fixed real number

0) = f (0,

1) = 0,

- and from (iii) we get f (0,
- x) = f (x,
- 0) = xk f (1,

Assume therefore that 0 < x ≤ y < 1

Notice that the condition (ii) gives the value for one class of pairs from G and that each pair in G can be reduced to one of the members of the class

This implies x = yk−1 x

- y) = f (y,
- x) = yk f 1,

y This can be written as f (x,

- y) = min(x,
- y)(max(u,
- v))k−1 for all 0 < x,

and the already obtained results we get 2 1 1 1 k−1 1 ,

- y = f x,
- f f f x,
- y = f x = f x,
- 2 2 2 2 1 Let us now consider x ≤ 2k−1 y in order to simplify the expression to the form f x,

yk−1 = 2 y k−1 k−1 2 x ,

and if we take x for which 2x ≤ y we get k − 1 = (k − 1) ,

- k = 1 or k = 2
- 2 For k = 1 the solution is f (x,
- y) = min(x,

and for k = 2 the solution is f (x,

- y) = xy

- △ Problem 23

(APMO 1989) Find all strictly increasing functions f : R → R such that f (x) + g(x) = 2x,

- where g is the inverse of f

Solution

Namely denote by Sd the the set of all numbers x for which f (x) = x + d

according to the definition of the inverse function we have g(x + d) = x,

and the given equation implies f (x + d) = x + 2d,

- x + d'∈ Sd

- where d'′ < d

From the above we have that each of those sets is infinite,

- if x belongs to some of them,

then each x + kd belongs to it as well

Let us use this to get the contradiction

- then y 6∈ Sd ′

which is a contradiction to our assumption

By further induction we prove that every y satisfying x + k(d − d'′) ≤ y < x + (k + 1)(d − d'′ ),

Marko Radovanovi´c: Functional Equations

- can’t be a member of Sd ′

However this is a contradiction with the previously established properties of the sets Sd and Sd ′

- △ Problem 24

hence it is not possible to form a recurrent equation

therefore h(h(1)) ≤ 2 and h(2) ≤ 2

Plugging n = 2 in the given equality gives 4 = h(h(2)) + h(3) = h(1) + h(3)

Let h(1) = k

- and that k ≤ 3

- hence h(3) = 1

This means that there are no solutions in this case

- 2◦ h(2) = 2

Then h(h(1)) = 1

From the equation for n = 2 we get h(3) = 2

Setting n = 3,

- 5 we get h(4) = 3,
- h(5) = 4,
- h(6) = 4,

and by induction we easily prove that h(n) ≥ 2,

- for n ≥ 2

Hence it is enough to guess some function and prove that it indeed solves the equation (induction or something similar sounds fine)

√ −1 + 5 where α = (this constant can be easily found α 2 + α = 1)

Proof that this is a 2 solution uses some properties of the integer part (although it is not completely trivial)

- △ Problem 25
- (IMO 2004,

shortlist) Find all functions f : R → R satisfying the equality f (x2 + y2 + 2 f (xy)) = f (x + y)2

Let us make the substitution z = x + y,

Given z,t ∈ R,

y are real if and only if 4t ≤ z2

Now the given functional equation transforms into f z2 + g(t) = ( f (z))2 for all t,

- z ∈ R with z2 ≥ 4t

f (z2 + c) = ( f (z))2 for all z ∈ R

Substituting t = 0 into (8) gives us

then taking z such that z2 + c'= 0,

we obtain from (9) that f (z)2 = c/2,

- which is impossible
- hence c'≥ 0

We also observe that x > c'implies f (x) ≥ 0

If g is a constant function,

we easily find that c'= 0 and therefore f (x) = x,

- which is indeed a solution

- and let a,

b ∈ R be such that g(a)− g(b) = d'> 0

For some sufficiently large K and each u,

v ≥ K with v2 − u2 = d'the equality u2 + g(a) = v2 + g(b) by (8) and (10) implies f (u) = f (v)

This further leads to g(u) − g(v) = 2(v − u) = √d 2

Therefore every value from u+

some suitably chosen segment [δ ,

with u and v bounded from above by some M

Olympiad Training Materials,

- imomath
- √ Consider any x,

y with y > x ≥ 2 M and δ < y2 − x2 < 2δ

By the above considerations,

- there exist u,

v ≤ M such that g(u) − g(v) = y2 − x2 ,

- x2 + g(u) = y2 + g(v)

Since x2 ≥ 4u and y2 ≥ 4v,

- 2 (8) leads to f (x)2 = f (y)2

- if we assume √ w
- that 4M ≥ c',

we conclude from (10) that f (x) = f (y)

Since this holds for any x,

y ≥√2 M with y2 − x2 ∈ [δ ,

it follows that f (x) is eventually constant,

say f (x) = k for x ≥ N = 2 M

Setting x > N in (9) we obtain k2 = k,

- so k = 0 or k = 1

and thus | f (z)| ≤ 1 for all z ≤ −N

which implies that g is unbounded

and consequently f (z)2 = f (z2 + g(t)) = k = k2

- then f (x) ≡ 0,
- which is clearly a solution

Then c'= 2 f (0) = 2 (because c'≥ 0),

which together with (10) implies f (x) = 1 for all x ≥ 2

Suppose that f (t) = −1 for some t < 2

then for some z ∈ R we have z2 = t − g(t) > 4t,

which by (8) leads to f (z)2 = f (z2 + g(t)) = f (t) = −1,

- which is impossible

- giving us t < −2/3

if X is any subset of (−∞,

- −2/3),

the function f defined by f (x) = −1 for x ∈ X and f (x) = 1 satisfies the requirements of the problem

To sum up,

- the solutions are f (x) = x,

f (x) = 0 and all functions of the form 1,

- x 6∈ X,
- f (x) = −1,
- x ∈ X,
- where X ⊂ (−∞,
- −2/3)

Problems for Independent Study

Most of the ideas for solving the problems below are already mentioned in the introduction or in the section with solved problems

Before solving the problems we highly encourage you to first solve (or look at the solutions) the problems from the previous section

Find all functions f : Q → Q that satisfy f (x + y) = f (x) + f (y) + xy

- for every natural number n

Find all functions f : N → N for which f (n) is a square of an integer for all n ∈ N,

and that satisfy f (m + n) = f (m) + f (n) + 2mn for all m,

- n ∈ N

Find all functions f : R → R that satisfy f ((x − y)2 ) = f (x)2 − 2x f (y) + y2

Let n ∈ N

Find all monotone functions f : R → R such that f (x + f (y)) = f (x) + yn

(USA 2002) Find all functions f : R → R which satisfy the equality f (x2 − y2 ) = x f (x) − y f (y)

- (Mathematical High Schol,

Belgrade 2004) Find all functions f : N → N such that f ( f (m) + f (n)) = m + n for every two natural numbers m and n

- (IMO 1983,

problem 1) Find all functions f : R → R such that (i) f (x f (y)) = y f (x),

- for all x,
- y ∈ R

(ii) f (x) → 0 as x → +∞

Let f : N → N be strictly increasing function that satisfies f ( f (n)) = 3n for every natural number n

- (IMO 1989,

shortlist) Let 0 < a < 1 be a real number and f continuous function on [0,

- 1] which satisfies f (0) = 0,
- f (1) = 1,

and x + y f = (1 − a) f (x) + a f (y),

- for every two real numbers x,
- y ∈ [0,
- 1] such that x ≤ y

- (IMO 1996,

shortlist) Let f : R → R be the function such that | f (x)| ≤ 1 and 13 1 1 f x+ + f (x) = f x + + f x+

- 42 6 7 Prove that f is periodic
- (BMO 2003,

problem 3) Find all functions f : Q → R that satisfy: (i) f (x + y) − y f (x) − x f (y) = f (x) f (y) − x − y + xy for every x,

- y ∈ Q

(ii) f (x) = 2 f (x + 1) + 2 + x,

- for every x ∈ Q
- (iii) f (1) + 1 > 0
- (IMO 1990,

problem 4) Determine the function f : Q+ → Q+ such that f (x f (y)) =

- f (x) ,
- for all x,
- y ∈ Q+
- (IMO 2002,

shortlist) Find all functions f : R → R such that f ( f (x) + y) = 2x + f ( f (y) − x)

(Iran 1997) Let f : R → R be an increasing function such that for all positive real numbers x and y: f (x + y) + f ( f (x) + f (y)) = f ( f (x + f (y)) + f (y + f (x)))

- (IMO 1992,

problem 2) Find all functions f : R → R,

such that f (x2 + f (y)) = y + f (x)2 for all x,

- y ∈ R
- (IMO 1994,

problem 5) Let S be the set of all real numbers strictly greater than

Find all functions f : S → S that satisfy the following two conditions: (i) f (x + f (y) + x f (y)) = y + f (x) + y f (x) for all x,

- y ∈ S

f (x) (ii) is strictly increasing on each of the intervals −1 < x < 0 and 0 < x

- (IMO 1994,

shortlist) Find all functions f : R+ → R such that f (x) f (y) = yα f (x/2) + xβ f (y/2),

- for all x,
- y ∈ R+
- (IMO 2002,

problem 5) Find all functions f : R → R such that ( f (x) + f (z))( f (y) + f (t)) = f (xy − zt) + f (xt + yz)

Olympiad Training Materials,

- imomath

(Vietnam 2005) Find all values for a real parameter α for which there exists exactly one function f : R → R satisfying f (x2 + y + f (y)) = f (x)2 + α · y

- (IMO 1998,

problem 3) Find the least possible value for f (1998) where f : N → N is a function that satisfies f (n2 f (m)) = m f (n)2

Does there exist a function f : N → N such that f ( f (n − 1)) = f (n + 1) − f (n) for each natural number n

- (IMO 1987,

problem 4) Does there exist a function f : N0 → N0 such that f ( f (n)) = n + 1987

Assume that the function f : N → N satisfies f (n + 1) > f ( f (n)),

- for every n ∈ N

that satisfy: (i) 2 f (m2 + n2 ) = f (m)2 + f (n)2 ,

for every two natural numbers m and n

(ii) If m ≥ n then f (m2 ) ≥ f (n2 )

(ii) f (mn) = f (m) f (n) for every two relatively prime natural numbers m and n

(iii) f (m) < f (n) whenever m < n

∞) that satisfy conditions (i) and (ii9) of the previous problem and the condition (ii) is modified to require the equality for every two natural numbers m and n

find all functions f : N0 → N0 for which f ( f (n)) + f (n) = 2n + 3k,

- for every n ∈ N0

(Vijetnam 2005) Find all functions f : R → R that satisfy f ( f (x − y)) = f (x) f (y) − f (x) + f (y) − xy

(China 1996) The function f : R → R satisfy f (x3 + y3 ) = (x + y) f (x)2 − f (x) f (y) + f (y)2 ,

- for all real numbers x and y

Find all functions f : R → R that satisfy: (i) f (x + y) = f (x) + f (y) for every two real numbers x and y

1 f (x) (ii) f = 2 for x 6= 0

- (IMO 1989,

shortlist) A function f : Q → R satisfy the following conditions: (i) f (0) = 0,

- f (α ) > 0 za α 6= 0

(ii) f (αβ ) = f (α ) f (β ) i f (α + β ) ≤ f (α ) + f (β ),

- for all α ,

β ∈ Q

(iii) f (m) ≤ 1989 za m ∈ Z

Prove that f (α + β ) = max{ f (α ),

f (β )} whenever f (α ) 6= f (β )

Find all functions f : R → R such that for every two real numbers x 6= y the equality f

- x + y x−y
- f (x) + f (y) f (x) − f (y)
- is satisfied

(ii) f (x3 ) = f (x)3 for all x ∈ Q+

Find all continuous functions f : R → R that satisfy the equality f (x + y) + f (xy) = f (x) + f (y) + f (xy + 1)

Find all continuous functions f ,

k : R → R that satisfy the equality f (x + y) + g(x − y) = 2h(x) + 2k(y)

- (IMO 1996,

shortlist) Find all functions f : N0 → N0 such that f (m + f (n)) = f ( f (m)) + f (n)

- (IMO 1995,

shortlist) Does there exist a function f : R → R satisfying the conditions: (i) There exists a positive real number M such that −M ≤ f (x) ≤ M for all x ∈ R

- (ii) f (1) = 1
- 2 1 1

? (iii) If x 6= 0 then f x + 2 = f (x) + f x x

(Belarus) Find all continuous functions f : R → R that satisfy f ( f (x)) = f (x) + 2x

- x + y 2
- 2xy = f (x) + f (y),

√ the it satisfy the equality 2 f ( xy) = f (x) + f (y) as well

Find all continuous functions f : (0,

- ∞) → (0,

∞) that satisfy f (x) f (y) = f (xy) + f (x/y)

for every two real numbers x and y

- imomath

(IMC 2001) Prove that there doesn’t exist a function f : R → R for which f (0) > 0 and f (x + y) ≥ f (x) + y f ( f (x))

(Romania 1998) Find all functions u : R → R for which there exists a strictly monotone function f : R → R such that f (x + y) = f (x)u(y) + f (y),

- y ∈ R

(Iran 1999) Find all functions f : R → R for which f ( f (x) + y) = f (x2 − y) + 4 f (x)y

- (IMO 1988,

problem 3) A function f : N → N satisfies: (i) f (1) = 1,

- f (3) = 3
- (ii) f (2n) = f (n)

(iii) f (4n + 1) = 2 f (2n + 1) − f (n) and f (4n + 3) = 3 f (2n + 1) − 2 f (n),

for every natural number n ∈ N

- (IMO 2000,

shortlist) Given a function F : N0 → N0 ,

assume that for n ≥ 0 the following relations hold: (i) F(4n) = F(2n) + F(n)

- (ii) F(4n + 2) = F(4n) + 1
- (iii) F(2n + 1) = F(2n) + 1

Prove that for every natural number m,

the number of positive integers n such that 0 ≤ n < 2m and F(4n) = F(3n) is equal to F(2m+1 )

- z) = f (x,
- z) f (y,
- xy) = f (z,
- x) f (z,
- 1 − x) = 1,
- for all rational numbers x,

Prove that f (x,

- x) = 1,
- −x) = 1,
- and f (x,
- y) f (y,

Find all functions f : N × N → R that satisfy f (x,

- x) = x,
- y) = f (y,
- (x + y) f (x,
- y) = y f (x,