Log In Register

Functional equations (Table of contents)

Functional Equations: Problems with SolutionsThe following problems are related to functional equations. Many of the problems were given at national andinternational mathematical competitions and olympiads, and thus are challenging. You may want to read first anintroductory text to Functional equations.

Problem 1 Find all functions such that and .

Hide solution.

This is a classical example of a problem that can be solved using mathematical induction. Notice that if weset and in the original equation we get , and since wehave for every natural number . Similarly for and we get

, i.e. . Now our goal is to find for each . Substituting and in the original equation gives us , and setting and gives

. Hence for each . Now we have to determine

. Plugging and we get

Furthermore for and we get , hence by the

mathematical induction . From (1) we now have

for every natural number . Furthermore for and we get , i.e.

, for every positive rational number . Setting and we get as well hence , for each .

Verification: Since , for all , is the solution toour equation.

Problem 2 (Belarus 1997) Find all functions such that for arbitrary real numbers and :

Hide solution.

Notice that and are obviously solutions to the given equation. Using mathematical

IMOmath Olympiads Book Training IMO Results Forum

f : Q → Q f(1) = 2 f(xy) = f(x)f(y) − f(x + y) + 1

x = 1 y = n f(n + 1) = f(n) + 1 f(1) = 2f(n) = n + 1 n x = 0 y = n

f(0)n = f(n) − 1 = n f(0) f(z) z ∈ Z x = −1y = 1 f(−1) = 0 x = −1 y = nf(−n) = −f(n − 1) + 1 = −n + 1 f(z) = z + 1 z ∈ Zf( )1

nx = n y = 1

n

f(1) = (n + 1)f( ) − f(n + ) + 1. (1)1n

1n

x = 1 y = m + 1n

f(m + 1 + ) = f(m + ) + 11n

1n

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

1n

f( ) = + 1,1n

1n

n x = m y = 1n

f( ) = + 1mn

mn

f(r) = r + 1 r x = −1 y = rf(−r) = −f(r − 1) + 1 = −r + 1 f(x) = x + 1 x ∈ Q

xy + 1 = (x + 1)(y + 1) − (x + y + 1) + 1 x, y ∈ Q f

g : R → R x y

g(x + y) + g(x)g(y) = g(xy) + g(x) + g(y).

g(x) = 0 g(x) = 2( ) =

induction it is not difficult to prove that if is not equal to one of these two functions then for all . It is also easy to prove that and , where is rational and

real number. Particularly from the second equation for we get , hence setting in the initial equation gives . This means that for . Now we use

the standard method of extending to . Assume that . Choose such that .Then

which is clearly a contradiction. Similarly from we get another contradiction. Thus we must have for every . \noindent It is easy to verify that all three functions satisfy the given functional

equation.

Problem 3 The function satisfies for every . Find all solutions of the equation

.

Hide solution.

The domain of this function is , so there isn t much hope that this can be solved using mathematicalinduction. Notice that and if then clearly . This means that thefunction is injective. Since , because of injectivity we must have ,implying . If there were another such that , injectivity would imply

and .

Problem 4 Find all injective functions that satisfy:

Hide solution.

Setting and first, and , afterwards we get

Let us emphasize that this is one standard idea if the expression on one side is symmetric with respect tothe variables while the expression on the other side is not. Now we have

. From here we conclude that implies and now the induction gives , for every

. Specially if then for all positive integers . The injectivity of givesthat at odd numbers (except 1) the function has to take odd values. Let be the smallest natural numbersuch that for some . We have for . Therefore thenumbers are mapped into . If for some , then for

, which is a contradiction. If for some such that then , which is a contradiction to the existence of such . Itfollows that the numbers are mapped into . Hence

. Thus the solution is and , for .\\ It is easy to verifythat the function satisfies the given conditions.

Problem 5 (BMO 1997, 2000) Solve the functional equation

Hide solution.

g g(x) = xx ∈ Q g(r + x) = r + g(x) g(rx) = rg(x) r x

r = −1 g(−x) = −g(x)y = −x g(x = g( ))2

x2 g(x) ≥ 0 x ≥ 0R g(x) < x r ∈ Q g(x) < r < x

r > g(x) = g(x − r) + r ≥ r,

g(x) > xg(x) = x x ∈ R

f : R → R x + f(x) = f(f(x)) x ∈ Rf(f(x)) = 0

R ′f(f(x)) − f(x) = x f(x) = f(y) x = y

f(f(0)) = f(0) + 0 = f(0) f(0) = 0f(f(0)) = 0 x f(f(x)) = 0 = f(f(0))

f(x) = f(0) x = 0

f : N → R

(a) f(f(m) + f(n)) = f(f(m)) + f(n), (b) f(1) = 2, f(2) = 4.

m = 1 n m = n n = 1

f(f(1) + f(n)) = f(f(1)) + f(n), f(f(n) + f(1)) = f(f(n)) + f(1).

f(f(n)) = f(n) − f(1) + f(f(1)) = f(n) − 2 + f(2) = f(n) + 2f(n) = m f(m) = m + 2 f(m + 2k) = m + 2k + 2k ≥ 0 f(1) = 2 f(2n) = 2n + 2 n f

pk f(k) = 2p + 1 f(2p + 2s + 1) = 2p + 2s + 3 s ≥ 0

3, 5, … , 2p − 1 1, 3, … , 2p + 1 f(t) = 1 tm = n = t 4 = f(2) = f(f(t) + f(t)) = f(f(t)) + f(t) = 3 t

f(t) = 3 f(3 + 2k) = 5 + 2k t3, 5, … , 2p − 1 5, 7, … , 2p + 1

f(3 + 2k) = 5 + 2k f(1) = 2 f(n) = n + 2 n ≥ 2

f(xf(x) + f(y)) = y + f(x , x, y ∈ R.)2

In probelms of this type it is usually easy to prove that the functions are injective or surjective, if thefunctions are injective/surjective. In this case for we get . Since the functionon the righthand side is surjective the same must hold for the function on the lefthand side. This impliesthe surjectivity of . Injectivity is also easy to establish. Now there exists such that andsubstitution and yields . For we get . Therefore

, i.e. . Replacing with gives

hence for every real number . Consider now the two cases: \noindent\emphFirst case . Plugging gives , and after taking squares

. Clearly in thiscase we have for every real . \noindent\emphSecond case . Plugging gives , and after taking squares

. Now weconclude for every real number . \noindent It is easy to verify that and

are indeed the solutions.

Problem 6 (IMO 1979, shortlist) Given a function , if for every two real numbers and the equality

holds, prove that for every two realnumbers and .

Hide solution.

This is a clasical example of the equation that solution is based on a careful choice of values that areplugged in a functional equation. Plugging in we get . Plugging in we get

. Plugging in we get and hence for all real

and . On the other hand, plugging in and we get .

Hence it follows that , i.e.,

Plugging in we get . Hence, and consequently for all reals. Now (1) reduces to .Plugging in and , we obtain for all nonzero reals and . Since

, it trivially holds that when one of and is .

Problem 7 Does there exist a function such that for every real number ?

Hide solution.

After some attempts we can see that none of the first three methods leads to a progress. Notice that thefunction of the righthand side has exactly fixed points and that the function has exactly 4 fixedpoints. Now we will prove that there is no function such that . Assume the contrary. Let bethe fixed points of , and the fixed points of . Assume that . Then

, hence and has to be on of the fixed points of . If then from we get a contradiction. Similarly , and since we get .Thus and . Furthermore we have . Let

. We immediately have , hence . Similarly if we get , and now we will prove that this is not possible. Take first

. Then which is clearly impossible. Similarly and (for otherwise ) hence . However we then have

, which is a contradiction, again. This proves that the required doesn texist.

x = 0 f(f(y)) = y + f(0)2

f t f(t) = 0x = 0 y = t f(0) = t + f(0)2

x = t f(f(y)) = y

t = f(f(t)) = f(0) = t + f(0)2f(0) = 0 x f(x)

f(f(x)x + f(y)) = + y,x2

f(x =)2x2 x

f(1) = 1 x = 1 f(1 + f(y)) = 1 + y

(1 + y = f(1 + f(y) = (1 + f(y) = 1 + 2f(y) + f(y = 1 + 2f(y) +)2 )2 )2 )2y2

f(y) = y y f(1) = −1 x = −1f(−1 + f(y)) = 1 + y

(1 + y = f(−1 + f(y) = (−1 + f(y) = 1 − 2f(y) + f(y = 1 − 2f(y) +)2 )2 )2 )2y2

f(y) = −y y f(x) = xf(x) = −x

f : R → R x yf(xy + x + y) = f(xy) + f(x) + f(y) f(x + y) = f(x) + f(y)

x y

x = y = 0 f(0) = 0 y = −1f(x) = −f(−x) y = 1 f(2x + 1) = 2f(x) + f(1)f(2(u + v + uv) + 1) = 2f(u + v + uv) + f(1) = 2f(uv) + 2f(u) + 2f(v) + f(1) u

v x = u y = 2v + 1f(2(u + v + uv) + 1) = f(u + (2v + 1) + u(2v + 1)) = f(u) + 2f(v) + f(1) + f(2uv + u)

2f(uv) + 2f(u) + 2f(v) + f(1) = f(u) + 2f(v) + f(1) + f(2uv + u)

f(2uv + u) = 2f(uv) + f(u). (1)

v = −1/2 0 = 2f(−u/2) + f(u) = −2f(u/2) + f(u) f(u) = 2f(u/2)f(2x) = 2f(x) f(2uv + u) = f(2uv) + f(u)

u = y x = 2uv f(x) + f(y) = f(x + y) x yf(0) = 0 f(x + y) = f(x) + f(y) x y 0

f : R → R f(f(x)) = − 2x2 x

g 2 g ∘ gf f ∘ f = g a, b

g a, b, c, d g ∘ g g(c) = yc = g(g(c)) = g(y) g(g(y)) = g(c) = y y g ∘ g y = a

a = g(a) = g(y) = c y ≠ b y ≠ c y = dg(c) = d g(d) = c g(f(x)) = f(f(f(x))) = f(g(x))

∈ a, bx0 f( ) = f(g( )) = g(f( ))x0 x0 x0 f( ) ∈ a, bx0∈ a, b, c, dx1 f( ) ∈ a, b, c, dx1

f(c) = a f(a) = f(f(c)) = g(c) = d f(c) ≠ bf(c) ≠ c g(c) = c f(c) = df(d) = f(f(c)) = g(c) = d f ′

Problem 8 Find all functions such that for every two positive realnumbers .

Hide solution.

Obviously is one solution to the problem. The idea is to find such that anduse this to determine . For every such that we can find such and from the given

condition we get . However this is a contradiction since we got that implies .One of the consequences is that . Assume that for some . From the given equationwe conclude that is nonincreasing (because ). Let us prove that is decreasing. In orderto do that it is enough to prove that , for each . Assume that for every (

). Substituting in the given equation we get the obvious contradiction. This means thatthe function is decreasing and hence it is injective. Again everything will revolve around the idea of gettingrid of . Notice that , therefore

i.e. . The injectivity of implies that

. If we plug we get

where , and according to our assumption .\\ It is easy to verify that , for

, and satisfy the equation.

Problem 9 (IMO 2000, shortlist) Find all pairs of functions and such that for every two realnumbers the following relation holds:

Hide solution.

Let us first solve the problem under the assumption that for some . Setting in the givenequation yields . Then the given equation becomes

, so setting we get ,where and . Hence is a linear function, and consequently isalso linear. If we now substitute and in the given equation and comparethe coefficients, we easily find that

Now we prove the existence of such that . If then putting in the givenequation we obtain , so we can take . Now assume that .By replacing by in the given equation we obtain

and, analogously, . The given functional equation for gives

, where . In particular, is injective and is surjective, so there exists such that . Now the above two relations yield

f : →R+ R+ f(x)f(yf(x)) = f(x + y)x, y

f(x) ≡ 1 y yf(x) = x + yf(x) x ≥ 0x

f(x)−1y

f(x) = 1 f(x) > 1 f(x) = 1f(x) ≤ 1 f(x) < 1 x

f f(yf(x)) ≤ 1 ff(x) < 1 x f(x) = 1 x ∈ (0, a)

a > 0 x = y = 2a

3

f(yf(x)) x + y > yf(x)

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

f(x) = f(f(yf(x))(x + y − yf(x))) f

x = f(yf(x))(x + y − yf(x)) f(x) = a

f(y) = ,1

1 + αz

α =1−f(a)

af(a)α > 0 f(x) = 1

1+αx

α ∈ R+ f(x) ≡ 1

f : R → R g : R → Rx, y

f(x + g(y)) = xf(y) − yf(x) + g(x).

g(α) = 0 α y = αg(x) = (α + 1)f(x) − xf(α)

f(x + g(y)) = (α + 1 − y)f(x) + (f(y) − f(α))x y = α + 1 f(x + n) = mxn = g(α + 1) m = f(α + 1) − f(α) f g

f(x) = ax + b g(x) = cx + d

f(x) = and g(x) = cx − , c ∈ R ∖ −1.cx − c2

1 + cc2

α g(α) = 0 f(0) = 0 y = 0f(x + g(0)) = g(x) α = −g(0) f(0) = b ≠ 0

x g(x)f(g(x) + g(y)) = g(x)f(y) − yf(g(x)) + g(g(x))f(g(x) + g(y)) = g(y)f(x) − xf(g(y)) + g(g(y)) x = 0f(g(y)) = a − by a = g(0) g f c ∈ R

f(c) = 0

g(x)f(y) − ay + g(g(x)) = g(y)f(x) − ax + g(g(y)). (1)

(1) ( ( )) = ( ) ( ) − + ( ( )) + = ( ) − + (1)

Plugging in we get . Now becomes . For we have ,whence

Note that , since is injective. From the surjectivity of it follows that is surjectiveas well, so it takes the value .

Problem 10 (IMO 1992, shortlist) Find all functions which satisfy

Hide solution.

This is a typical example of a problem that is solved using recurrent equations. Let us define inductivelyas , where is a fixed real number. It follows from the given equation in that

. The general solution to this equation is of the form

where satisfy and . In order to have for all we must have . Hence and . Since was arbitrary, weconclude that is the only possible solution of the functional equation. It is easily verified that thisis indeed a solution.

Problem 11 (Vietnam 2003) Let be the set of all functions which satisfy the inequality , forevery positive real number . Find the largest real number such that for all functions :

.

Hide solution.

We clearly have that , hence . Furthermore for every function we have .The idea is the following: Denote and form a sequence for which and whichwill (hopefully) tend to . This would imply that , and hence . Let us constract a recurrentrelation for . Assume that , for every . From the given inequality we have

This means that . Let us prove that . This is a standard problem. It iseasy to prove that the sequence is increasing and bounded above by . Hence it converges and its

limit satisfies , i.e. (since ).

Problem 12 Find all functions that satisfy

Hide solution.

Our first goal is to express and using and get the equation involving only. First taking and

y = c (1) g(g(x)) = g(c)f(x) − ax + g(g(c)) + ac = kf(x) − ax + d (1)g(x)f(y) + kf(x) = g(y)f(x) + kf(y) y = 0 g(x)b + kf(x) = af(x) + kb

g(x) = f(x) + k.a − k

b

g(0) = a ≠ k = g(c) g f g0

f : →R+ R+

f(f(x)) + af(x) = b(a + b)x.

xn

= f( )xn xn−1 ≥ 0x0 f= −a + b(a + b)xn+2 xn+1 xn

= + (−a − b ,xn λ1bn λ2 )n

, ∈ Rλ1 λ2 = +x0 λ1 λ2 = b − (a + b)x1 λ1 λ2 ≥ 0xn n= 0λ2 =x0 λ1 f( ) = = b = bx0 x1 λ1 x0 x0

f(x) = bx

F f : →R+ R+ f(3x) ≥ f(f(2x)) + xx α f ∈ F

f(x) ≥ α ⋅ x

∈ Fx

2α ≤ 1

2f ∈ F f(x) ≥ x

3=1

3α1 αn f(x) ≥ xαn

12

α ≥ 12

α = 12

αk f(x) ≥ xαk x ∈ R+

f(3x) ≥ f(f(2x)) + x ≥ f(2x) + x ≥ ⋅ ⋅ 2x + x = ⋅ 3x.αk αk αk αk+1

=αn+12 +1α2

n

3=limn→+∞ αn

12

αk12

α α = 2 +1α2

3α = 1

2α < 1

f, g, h : R → R

f(x + y) + g(x − y) = 2h(x) + 2h(y).

f g h h y = x

(0) = (2 ) = 4 ( ) − .

substituting we get Furthermore by putting we get

, where . Now the original equation can be written as

Let . These "longer" linear expressions can be easily handled if we express functions inform of the sum of an even and odd function, i.e. . Substituting this into (1) andwriting the same expressions for and we can add them together and get:

If we set in this expression and add to (2) we get (using )

The last equation is not very difficult. Mathematical induction yields , for every rationalnumber . From the continuity we get . Similar method gives the simple relation for

This is a Cauchy equation hence . Thus and substituting for and we get:

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

Problem 13 Find all functions for which

Solve the same problem for the case .

Hide solution.

It is not hard to see that for we get , i.e. . Furthermore, setting and gives , hence or . We will separate this

into two cases:

Let . In this innocentlooking problems that are resistent to usual ideas it issometimes successful to increase the number of variables, i.e. to set instead of :

Although it seems that the situation is worse and running out of control, that is not the case. Namelythe expression on the lefthand side is symmetric, while the one on the righthand side is not. Writingthe same expression for and equating gives

Setting (we couldn t do that at the beginning, since was fixed) we get , and setting in this equality gives

Setting gives , i.e. or . This means that we have twocases here as well:

If , then from (2) plugging instead of we get . Setting

g(0) = a f(2x) = 4h(x) − a. y = 0g(x) = 2h(x) + 2b − 4h( ) + ax

2h(0) = b

2[h( ) + h( )] + h(x − y) + b = h(x) + h(y). (1)x + y

2x − y

2

H(x) = h(x) − bH(x) = (x) + (x)He Ho

(−x, y) (x, −y)

2[ ( ) + ( )] + (x − y) = (x) + (y). (2)He

x − y

2He

x + y

2He He He

−y (y) = (−y)He He

(x + y) − (x − y) = 2 (x) + 2 (y).He He He He

(r) = αHe r2

r (x) = αHe x2 Ho

(x + y) + (x − y) = 2 (x).Ho Ho Ho

(x) = βxHo h(x) = α + βx + bx2 f g

f(x) = α + 2βx + 4b − a, g(x) = α + a.x2 x2

f : Q → Q

f(xy) = f(x)f(y) − f(x + y) + 1.

f : R → R

x = y = 0 (f(0) − 1 = 0)2f(0) = 1

x = 1 y = −1 f(−1) = f(1)f(−1) f(−1) = 0 f(1) = 1

1∘ f(−1) = 0yz 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.

x

f(x)f(y + z) − f(x) + f(x + yz) = f(z)f(x + y) − f(z) + f(xy + z). (1)

z = −1 ′ z = 1f(x)f(z − 1) − f(x) + f(x − y) = f(xy − 1) x = 1

f(y − 1)(1 − f(1)) = f(1 − y) − f(1). (2)

y = 2 f(1)(2 − f(1)) = 0 f(1) = 0 f(1) = 2

1.1∘ f(1) = 0 y + 1 y f(y) = f(−y)

( ) = ( ) ( ) − ( − ) + 1

instead of in the initial equality gives , hence , for every two rational numbers and . Specially for we get

, for all . However this is a contradiction with . In this casewe don t have a solution.

If , setting instead of in (2) gives . It is clearthat we should do the substitution because the previous equality gives

, i.e. is odd. Furthermore substituting into the original equality gives

Setting instead of we get , and addingwith ( ) yields . For we have therefore we get . This is a the Cauchy equation and since thedomain is we get for some rational number . Plugging this back to (3) weobtain , and easy verification shows that satisfies the conditions of theproblem.

Let . Setting in (1) we get

hence for we get , for every rational . This means that and thisfunction satisfies the given equation.

Now let us solve the problem where . Notice that we haven t used that the range is , hencewe conclude that for all rational numbers , or . If for all rationalnumbers , it can be easily shown that . Assume that . From the above we have that

, hence it is enough to prove monotonicity. Substitute in (3) and use to get . Therefore for every positive the value is nonpositive.

Hence if , i.e. we have , and the function is decreasing.This means that and after some calculation we get . It is easy to verify thatso obtained functions satisfy the given functional equation.

Problem 14 (IMO 2003, shortlist) Let denote the set of positive real numbers. Find all functions that satisfy the following conditions:

(i)

(ii) for all .

Hide solution.

First notice that the solution of this functional equation is not one of the common solutions that we are usedto work with. Namely one of the solutions is which tells us that this equality is unlikely to beshown reducing to the Cauchy equation. First, setting we get (since ).One of the properties of the solution suggested above is , and proving this equality will beour next step. Putting , , in (i) gives

In particular, for the last equality yields ; hence for each . Itfollows that there exists such that . Now it follows by induction from (1) that

for every integer , and therefore for every rational . Consequently, if is fixed, we have , where . But since the set of ( ) is dense in

−y y f(xy) = f(x)f(y) − f(x − y) + 1f(x + y) = f(x − y) x y x = yf(2x) = f(0) = 1 x ∈ Q f(1) = 0

′

1.2∘ f(1) = 2 y + 1 y 1 − f(y) = f(−y) − 1g(x) = 1 − f(x)

g(−x) = −g(x) g g

g(xy) = g(x) + g(y) − g(x)g(y) − g(x + y). (3)

−y y −g(xy) = g(x) − g(y) + g(x)g(y) − g(x − y)?? g(x + y) + g(x − y) = 2g(x) x = y g(2x) = 2g(x)

g(x + y) + g(x − y) = g(2x)Q g(x) = rx r

r = −1 f(x) = 1 + x

2∘ f(1) = 1 z = 1

f(xy + 1) − f(x)f(y + 1) + f(x) = 1,

y = −1 f(1 − x) = 1 x f(x) ≡ 1

f : R → R ′ Qq f(q) = q + 1 f(q) ≡ 1 f(q) = 1

q f(x) ≡ 1 f(q) ≡ 1g(x) + g(y) = g(x + y) x = y

g(2x) = 2g(x) g( ) = −g(xx2 )2r g(r)

y > x y = x + r2 g(y) = g(x) + g( ) ≤ g(x)r2

f(x) = 1 + αx f(x) = 1 + x

R+ f : →R+ R+

f(xyz) + f(x) + f(y) + f(z) = f( )f( )f( )xy−−√ yz−−√ zx−−√

f(x) < f(y) 1 ≤ x < y

f(x) = x + 1x

x = y = z = 1 f(1) = 2 f(1) > 0f(x) = f(1/x)

x = ts y = ts

z = st

f(t)f(s) = f(ts) + f(t/s). (1)

s = 1 f(t) = f(1/t) f(t) ≥ f(1) = 2 t

g(t) ≥ 1 f(t) = g(t) + 1g(t)

g( ) = g(ttn )nn g( ) = g(ttq )q

qt > 1 f( ) = +tq aq a−q a = g(t) aq q ∈ Q

+

and is monotone on and , it follows that for every real .Therefore, if is such that , we have

Problem 15 Find all functions that satisfy:

(i) for every ;

(ii) for every .

Hide solution.

It is not hard to see that is a solution. Let us prove that this is the only solution. Using thegiven conditions we get

i.e. . With this we have found the upper bound for . Since our goal is to prove we will use the same method for lowering the upper bound. Similarly we get

Now it is clear that we should use induction to prove

for every . However this is shown in the same way as the previous two inequalities. Since as , hence for fixed we can t have . This implies for every real

number . It remains to show that , for . We will use the similar argument. From

the fact that the range is we get , i.e. . Wefurther have and similarly by induction

Passing to the limit we further have . Now again from the given equality we get , i.el . Using the induction we get

, and passing to the limit we get the required inequality .

Problem 16 (IMO 1999, probelm 6) Find all functions such that

Hide solution.

Let , i.e. . We will determine the value of the function on . Let , for some . From the given equality we have , i.e.

where . Now it is clear that we have to analyze set further. Setting in the original

R+ f (0, 1] [1, ∞) f( ) = +tr ar a−r r

k = atk

f(x) = + for every x ∈ R.xk x−k

f : [1, ∞) → [1, ∞)

f(x) ≤ 2(1 + x) x ∈ [1, ∞)

xf(x + 1) = f(x − 1)2x ∈ [1, ∞)

f(x) = x + 1

f(x = xf(x + 1) + 1 ≤ x(2(x + 1)) + 1 < 2(1 + x ,)2 )2

f(x) ≤ (1 + x)2√ f(x)f(x) = x + 1

f(x = xf(x + 1) + 1 ≤ x( (x + 1)) + 1 < (1 + x .)2 2√ 21/4 )2

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

k → 121/2k

k → +∞ x ′ f(x) > x + 1 f(x) ≤ x + 1x ≥ 1 f(x) ≥ x + 1 x ≥ 1

[1, +∞) = f(x + 1) ≥ 1f(x −1)2

xf(x) ≥ >x + 1− −−−−√ x1/2

f(x = 1 + xf(x + 1) > 1 + x >)2x + 2− −−−−√ x3/2

f(x) > .x1−1/2k

f(x) ≥ x

f(x = 1 + xf(x + 1) ≥ (x + 1/2)2 )2f(x) ≥ x + 1/2

f(x) ≥ x + 1 − 12k

f(x) ≥ x + 1

f : R → R

f(x − f(y)) = f(f(y)) + xf(y) + f(x) − 1.

A = f(x) | x ∈ R A = f(R) Ax = f(y) ∈ A y f(0) = f(x) + + f(x) − 1x2

f(x) = − ,c + 1

2x2

2

f(0) = c A x = y = 0

(− ) = ( ) + − 1

equation we get , hence . Furthermore, plugging in the originalequation we get . Since the range of the function (on ) on the righthand side is entire , we get , i.e. . Hence for every realnumber there are real numbers such that . Now we have

From the original equation we easily get . It is easy to show that the function satisfies the given equation.

Problem 17 Given an integer , let be a continuous function satisfying , , and

, for every . Prove that for each .

Hide solution.

First from we have , hence is injective. The idea for what follows isclear once we look at the graphical representation. Namely from the picture it can be easily deduced thatthe function has to be strictly increasing. Let us prove that formally. Assume the contrary, that for some tworeal numbers we have . The continuity on implies that there is some such that , which contradicts the injectivity of . Now if , we get etc. . Similarly we get a contradiction if we assume that . Hence for each

we must have .

Problem 18 Find all functions that satisfy for all

.

Hide solution.

Clearly is one solution to the functional equation. Let us prove that the function is nonincreasing. Assume the contrary that for some we have . We will consider

the expression of the form since it is positive and bigger then . We first plug instead of in the original equation, then we plug instead of , we get

, which is a contradiction. Hence the function is nondecreasing.

Let us prove that . Let . Substituting we get , hence for . Therefore the function is periodic on the interval , and

since it is monotone it is constant. However we then conclude that the lefthand side of the originalequation constant and the righthand side is not. Thus we must have . Let us prove that

for . Indeed for the given equality gives . If

we have and . If we have

, and . Hence . If , plugging we get

and since , we get , i.e. in this case, too. This means that forall positive real numbers .

Problem 19 (Bulgaria 1998)

f(−c) = f(c) + c − 1 c ≠ 0 y = 0f(x − c) − f(x) = cx + f(c) − 1 x

R f(x − c) − f(x) | x ∈ R = R A − A = Rx , ∈ Ay1 y2 x = −y1 y2

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

= f( ) + f( ) + − 1 = c − .y1 y2 y1y2x2

2

c = 1 f(x) = 1 − x2

2

n f : R → R f(0) = 0 f(1) = 1(x) = xf (n) x ∈ [0, 1] f(x) = x x ∈ [0, 1]

f(x) = f(y) (x) = (y)f (n) f (n) f

<x1 x2 f( ) ≥ f( )x1 x2 [0, ]x1 cf(c) = f( )x2 f x < f(x) f(x) < f(f(x))

x < f(n)(x) = x x > f(x)x ∈ [0, 1] f(x) = x

f : (0, +∞) → (0, +∞) f(f(x) + y) = xf(1 + xy)x, y ∈ (0, +∞)

f(x) = 1x

0 < x < y 0 < f(x) < f(y)z =

yf(y)−xf(x)y−x

f(y)(x, z − f(y)) (x, y) z − f(x) yx = y

f(1) = 1 f(1) ≠ 1 x = 1 f(f(1) + y) = f(1 + y)f(u + |f(1) − 1|) = f(u) u > 1 (1, +∞)

f(1) = 1f(x) = 1

xx > 1 y = 1 − 1

xf(f(x) − ) = xf(x)1

x

f(x) > 1x

f(f(x) − + 1) ≤ f(1) = 11x

xf(x) > 1 f(x) < 1x

f(f(x) − + 1) ≥ f(1) = 11x

xf(x) < 1 f(x) = 1x

x < 1 y = 1x

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

x

2

≥ 11x

f(x) + =1x

2x

f(x) = 1x

f(x) = 1x

x

Prove that there is no function such that for every twopositive real numbers and .

Hide solution.

The common idea for the problems of this type is to prove that for some which will leadus to the obvious contradiction. We can also see that it is sufficient to prove that

, for every because the simple addition gives . Forsufficiently large this implies . Hence our goal is finding such that

, for every . Assume that such function exists. From the given inequality we get

and the function is obviously decreasing. Also from the given equality we can

conclude that

Let be a natural number such that (such number clearly exists). Notice that for the following inequality holds

and adding similar realitions for all described yields which is a contradiction.

Problem 20 Let be a function satisfying

Find the number of integers for which .

Hide solution.

This is a typical problem in which the numbers should be considered in some base different than 10. Forthis situation the base is doing the job. Let us calculate for in an attempt to guess thesolution. Clearly the given equation can have only one solution.

Now we see that is obtained from by changing each digit by , and conversely. This can be noweasily shown by induction. It is clear that if and only if in the system with base 3 doesn tcontain any digit (because this would imply ). Now it is easy to count the number of such s.The answer is 127.

Problem 21 (BMO 2003, shortlist) Find all possible values for if is the function satisfying the conditions:

(i) for all ;

(ii) for all ;

(iii) .

f : →R+ R+ f(x ≥ f(x + y)(f(x) + y))2

x y

f(y) < 0 y > 0

f(x) − f(x + 1) ≥ c > 0 x f(x) − f(x + m) ≥ mcm f(x + m) < 0 c

f(x) − f(x + 1) ≥ c x

f(x) − f(x + y) ≥f(x+y)y

f(x)

f(x) − f(x + y) ≥ .f(x)y

f(x) + y

n f(x + 1)n ≥ 10 ≤ k ≤ n − 1

f(x + ) − f(x + ) ≥ ≥ ,k

n

k + 1n

f(x + )kn

1n

f(x + ) +kn

1n

12n

k f(x) − f(x + 1) ≥ 12

f : N → N

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

n ≤ 2006 f(n) = 2n

3 f(n) n ≤ 8

f((1 ) = (2 , f((2 ) = (1 , f((10 ) = 6 = (20 , f((11 ) = 8 = (22 ,)3 )3 )3 )3 )3 )3 )3 )3

f((12 ) = 7 = (21 , f((20 ) = 3 = (10 , f((21 ) = 5 = (12 , f((22 ) = 4 = (11 .)3 )3 )3 )3 )3 )3 )3 )3

f(n) n 2 1f(n) = 2n n ′

1 f(n) < 2n n′

f( )20042003

f : Q → [0, +∞)

f(xy) = f(x)f(y) x, y ∈ Q

f(x) ≤ 1 ⇒ f(x + 1) ≤ 1 x ∈ Q

f( ) = 220032002

Hide solution.

Notice that from (i) and (ii) we conclude that , for every rational . Now (i) implies that for we get and similarly for we get . By induction

for every integer . For from we have that , and according

to (ii) This implies

hence , for every . Now you might wonder how did we get thisidea. There is one often neglected fact that for every two relatively prime numbers and , there areintegers and such that . What is all of this good for? We got that , and we knowthat for all and since is the maximum of the function on and since we have theprevious inequality our goal is to show that the value of the function is for a bigger class of integers. Wewill do this for prime numbers. If for every prime we have then for every integerimplying which contradicts (iii). Assume therefore that for some . There are and such that implying Now we must have

implying that for every other prime number . From (iii) we have

hence only one of the numbers is equal to . Thus giving:

If then , otherwise it is .

It remains to construct one function for each of the given values. For the first value it is the multiplicativefunction taking the value at the point , and for all other prime numbers; in the second case it is athe multiplicative function that takes the value at, for example, and takes at all other primenumbers. For these functions we only need to verify the condition (ii), but that is also very easy to verify.

Problem 22 Let , and . Find all such that for all the followingstatements hold:

(i) ;

(ii) , ;

(iii) for every , where is a fixed real number.

Hide solution.

The function of several variables appears in this problem. In most cases we use the same methods as inthe case of a singlevariable functions. From the condition (ii) we get , and from (iii)we get . This means that is entirely defined on the edge of theregion . Assume therefore that . Notice that the condition (ii) gives the value for oneclass of pairs from and that each pair in can be reduced to one of the members of the class. Thisimplies

f(x) > 0 xx = y = 1 f(1) = 0 x = y = −1 f(−1) = 1 f(x) ≤ 1

x f(x) ≤ f(y) f( )f(y) = f(x)y

xf( ) ≤ 1

y

x

f( + 1) ≤ 1.y

x

f(x + y) = f( + 1)f(x) ≤ f(x),y

x

f(x + y) ≤ maxf(x), f(y) x, y ∈ Qu v

a b au + bv = 1 f(1) = 1f(x) ≤ 1 x ∈ Z 1 Z

1p f(p) = 1 f(x) = 1

f(x) ≡ 1 f(p) ≠ 1 p ∈ P ab ap + bq = 1 f(1) = f(ap + bq) ≤ maxf(ap), f(bq).

f(bq) = 1 f(q) = 1 q

f( ) = = 2,20032002

f(2003)f(2)f(7)f(11)f(13)

f(2), f(7), f(11), f(13) 1/2f(3) = f(167) = f(2003)

f( ) = = f(2 .20042003

f(2 f(3)f(167))2

f(2003))2

f(2) = 1/2 f( ) =20032002

14

1

1/2 2 11/2 7 1

I = [0, 1] G = I × I k ∈ N f : G → I x, y, z ∈ I

f(f(x, y), z) = f(x, f(y, z))

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

f(zx, zy) = f(x, y)zk x, y, z ∈ I k

f(1, 0) = f(0, 1) = 0f(0, x) = f(x, 0) = f(1, 0) = 0xk fG 0 < x ≤ y < 1

G G

( )

This can be written as for all . Let us find all possiblevalues for . Let . From the condition (i), and the already obtained results we get

Let us now consider in order to simplify the expression to the form

, and if we take for which we get , i.e.

or . For the solution is , and for the solution is . It is easy to verify that both solutions satisfy the given conditions.

Problem 23 (APMO 1989) Find all strictly increasing functions such that

where is the inverse of .

Hide solution.

Clearly every function of the form is the solution of the given equation. Another useful idea appearsin this problem. Namely denote by the the set of all numbers for which . Our goal is toprove that . Assume that is nonempty. Let us prove that for we have aswell. Since , according to the definition of the inverse function we have , andthe given equation implies , i.e. . Let us prove that the sets areempty, where . From the above we have that each of those sets is infinite, i.e. if belongs to someof them, then each belongs to it as well. Let us use this to get the contradiction. More precisely wewant to prove that if and , then . Assume the contrary. From themonotonicity we have , which is a contradiction to our assumption. Byfurther induction we prove that every satisfying

can t be a member of . However this is a contradiction with the previously established properties of thesets and . Similarly if switching the roles of and gives a contradiction. Simpleverification shows that each satisfies the given functional equation.

Problem 24 Find all functions that satisfy

Hide solution.

Notice that we have both and , hence it is not possible to form a recurrent equation. Wehave to use another approach to this problem. Let us first calculate and . Setting gives

, therefore and . Let us consider the two cases:

. Then . Plugging in the given equality gives. Let . It is clear that and , and that

. This means that , hence . However from we geta contradiction. This means that there are no solutions in this case.

. Then . From the equation for we get . Setting

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

yyk−1

f(x, y) = min(x, y)(max(u, v))k−1 0 < x, y < 1k 0 < x ≤ ≤ y < 11

2

f(f(x, ), y) = f(x( , y) = f(x, f( )) = f(x, ).12

12

)k−1 12

12

yk−1

x ≤ y2k−1

f(x, ) = x(12

yk−1 y

2)k−1

x 2x ≤ yk−1 k − 1 = (k − 1)2

k = 1 k = 2 k = 1 f(x, y) = min(x, y) k = 2f(x, y) = xy

f : R → R

f(x) + g(x) = 2x,

g f

x + dSd x f(x) = x + d

= RSd Sd x ∈ Sd x + d ∈ Sd

f(x) = x + d g(x + d) = xf(x + d) = x + 2d x + d ∈ Sd Sd ′

< dd ′ xx + kdx ∈ Sd x ≤ y ≤ x + (d − )d ′ y ∉ Sd ′

y + = f(y) ≥ f(x) = x + dd ′

y

x + k(d − ) ≤ y < x + (k + 1)(d − ),d ′ d ′

′ Sd ′

Sd Sd ′ > dd ′ d d ′

f(x) = x + d

h : N → N

h(h(n)) + h(n + 1) = n + 2.

h(h(n)) h(n + 1)h(1) h(2) n = 1

h(h(1)) + h(2) = 3 h(h(1)) ≤ 2 h(2) ≤ 2

1∘ h(2) = 1 h(h(1)) = 2 n = 24 = h(h(2)) + h(3) = h(1) + h(3) h(1) = k k ≠ 1 k ≠ 2k ≤ 3 k = 3 h(3) = 1 2 = h(h(1)) = h(3) = 1

2∘ h(2) = 2 h(h(1)) = 1 n = 2 h(3) = 2

(4) = 3, (5) = 4, (6) = 4

we get , and by induction we easily prove that, for . This means that . Clearly there is at most one function satisfying the

given equality. Hence it is enough to guess some function and prove that it indeed solves theequation (induction or something similar sounds fine). The solution is

where (this constant can be easily found ). Proof that this is a solutionuses some properties of the integer part (although it is not completely trivial).

Problem 25 (IMO 2004, shortlist) Find all functions satisfying the equality

Hide solution.

Let us make the substitution , . Given , are real if and only if .Define . Now the given functional equation transforms into

Let us set . Substituting into (1) gives us

If , then taking such that , we obtain from (2) that , which is impossible;hence . We also observe that

If is a constant function, we easily find that and therefore , which is indeed a solution.

Suppose is nonconstant, and let be such that . For some sufficientlylarge and each with the equality by (1) and (2)implies . This further leads to . Therefore every

value from some suitably chosen segment can be expressed as , with and bounded from above by some .

Consider any with and . By the above considerations, there exist such that , i.e., . Since and

, (1) leads to . Moreover, if we assume w.l.o.g. that , we conclude from(3) that . Since this holds for any with , it follows that is eventually constant, say for . Setting in (2) we obtain , so

or .

By (1) we have , and thus for all . Hence for , which implies that is unbounded. Hence for each

there exists such that , and consequently . Therefore for each .

If , then , which is clearly a solution. Assume . Then (because ), which together with (3) implies for all . Suppose that for some .

Then . If also , then for some we have ,which by (1) leads to , which is impossible. Hence ,giving us . On the other hand, if is any subset of , the function defined by

for and satisfies the requirements of the problem.

n = 3, 4, 5 h(4) = 3, h(5) = 4, h(6) = 4h(n) ≥ 2 n ≥ 2 h(1) = 1

h(n) = ⌊nα⌋ + 1,

α = −1+ 5√2

+ α = 1α2

f : R → R

f( + + 2f(xy)) = f(x + y .x2 y2 )2

z = x + y t = xy z, t ∈ R x, y 4t ≤ z2

g(x) = 2(f(x) − x)

f( + g(t)) = for all t, z ∈ R with ≥ 4t. (1)z2 (f(z))2z2

c = g(0) = 2f(0) t = 0

f( + c) = for all z ∈ R. (2)z2 (f(z))2

c < 0 z + c = 0z2 f(z = c/2)2

c ≥ 0

x > c implies f(x) ≥ 0. (3)

g c = 0 f(x) = x

g a, b ∈ R g(a) − g(b) = d > 0K u, v ≥ K − = dv2 u2 + g(a) = + g(b)u2 v2

f(u) = f(v) g(u) − g(v) = 2(v − u) = d

u+ +du2√

[δ, 2δ] g(u) − g(v) u vM

x, y y > x ≥ 2 M−−√ δ < − < 2δy2 x2

u, v ≤ M g(u) − g(v) = −y2 x2 + g(u) = + g(v)x2 y2 ≥ 4ux2

≥ 4vy2 f(x = f(y)2 )2 4M ≥ c2

f(x) = f(y) x, y ≥ 2 M−−√ − ∈ [δ, 2δ]y2 x2 f(x)

f(x) = k x ≥ N = 2 M−−√ x > N = kk2

k = 0 k = 1

f(−z) = ±f(z) |f(z)| ≤ 1 z ≤ −Ng(u) = 2f(u) − 2u ≥ −2 − 2u u ≤ −N g z

t + g(t) > Nz2 f(z = f( + g(t)) = k =)2z2 k2

f(z) = ±k z

k = 0 f(x) ≡ 0 k = 1 c = 2f(0) = 2c ≥ 0 f(x) = 1 x ≥ 2 f(t) = −1 t < 2

t − g(t) = 3t + 2 > 4t t − g(t) ≥ 0 z ∈ R = t − g(t) > 4tz2

f(z = f( + g(t)) = f(t) = −1)2z2 t − g(t) < 0

t < −2/3 X (−∞, −2/3) ff(x) = −1 x ∈ X f(x) = 1

( ) = ( ) = 0

To sum up, the solutions are , and all functions of the form

where .

Functional equations (Table of contents)

20052015 IMOmath.com | imomath"at"gmail.com | Math rendered by MathJaxHome | Olympiads | Book | Training | IMO Results | Forum | Links | About | Contact us

f(x) = x f(x) = 0

f(x) = 1,−1,

x ∉ X,x ∈ X,

X ⊂ (−∞, −2/3)