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

Functional Equations
Marko Radovanovi´
c
radmarko@yahoo
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Cauchy Equation and Equations of the Cauchy type
Problems with Solutions
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2

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14

Basic Methods For Solving Functional Equations
• Substituting the values for variables
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0 or 1), after that (if possible) some expressions which will make some part of the equation
to become constant
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Substitutions become less obvious as the difficulty of the problems
increase
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This method relies on using the value f (1) to find all f (n) for n
1
integer
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This method is used in problems
n
where the function is defined on Q and is very useful, especially with easier problems
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In many of
the problems these facts are not difficult to establish but can be of great importance
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The number of problems using this method is
considerably smaller than the number of problems using some of the previous three methods
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• Using the Cauchy’s equation and equation of its type
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Continuity is usually given as
additional condition and as the monotonicity it usually serves for reducing the problem to
Cauchy’s equation
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• 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
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• Making recurrent relations
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Olympiad Training Materials, www
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imocompendium
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The goal
is to prove that the described set is precisely the domain of the function
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This method is often used to simplify the given equation and is
seldom of crucial importance
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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
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Of course, this can be used only if
the domain is N
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This can help a lot in finding the appropriate substitutions
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2

Cauchy Equation and Equations of the Cauchy type

The equation f (x + y) = f (x) + f (y) is called the Cauchy equation
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That fact is easy to prove using mathematical
induction
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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) = xf (1)
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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) = xf (1)
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• All continuous functions f : R → (0, +∞) satisfying f (x + y) = f (x)f (y) are of the form
f (x) = ax
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• All continuous functions f : (0, +∞) → R satisfying f (xy) = f (x) + f (y) are of the
form f (x) = loga x
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• 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
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3

Problems with Solutions

The following examples should illustrate the previously outlined methods
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Find all functions f : Q → Q such that f (1) = 2 and f (xy) = f (x)f (y)−f (x+y)+1
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This is a classical example of a problem that can be solved using mathematical induction
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Similarly for x = 0 and y = n
we get f (0)n = f (n) − 1 = n, i
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f (0)
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Substituting
x = −1 and y = 1 in the original equation gives us f (−1) = 0, and setting x = −1 and y = n
gives f (−n) = −f (n − 1) + 1 = −n + 1
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Now we have to
1
1

...

n
n

(1)

1
1
1
= f m+
+ 1, hence by the
Furthermore for x = 1 and y = m + we get f m + 1 +
n
n
n
1
1
mathematical induction f m +
=m+f

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e
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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
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△
for every natural number n
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(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)
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Notice that g(x) = 0 and g(x) = 2 are obviously solutions to the given equation
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It is also easy to prove that g(r + x) = r + g(x) and g(rx) = rg(x),
where r is rational and x real number
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This means that
g(x) ≥ 0 for x ≥ 0
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Assume that g(x) < x
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Then
r > g(x) = g(x − r) + r ≥ r,
which is clearly a contradiction
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Thus we
must have g(x) = x for every x ∈ R
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△
Problem 3
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Find all
solutions of the equation f (f (x)) = 0
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The domain of this function is R, so there isn’t much hope that this can be solved using
mathematical induction
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This means that the function is injective
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If there were another x such that f (f (x)) = 0 =
f (f (0)), injectivity would imply f (x) = f (0) and x = 0
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Find all injective functions f : N → R that satisfy:
(a) f (f (m) + f (n)) = f (f (m)) + f (n),

(b) f (1) = 2, f (2) = 4
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imo
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yu, www
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com

Solution
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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
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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
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The injectivity of f gives
that at odd numbers (except 1) the function has to take odd values
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We have f (2p + 2s + 1) = 2p + 2s + 3 for s ≥ 0
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, 2p − 1 are mapped into 1, 3,
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If f (t) = 1 for some t,
then for m = n = t 4 = f (2) = f (f (t) + f (t)) = f (f (t)) + f (t) = 3, which is a contradiction
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It follows that the numbers 3, 5,
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, 2p + 1
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Thus the solution is f (1) = 2 and f (n) = n + 2, for n ≥ 2
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△
Problem 5
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Solution
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In this case for x = 0 we get f (f (y)) = y+f (0)2
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This implies the surjectivity of f
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Now there exists
t such that f (t) = 0 and substitution x = 0 and y = t yields f (0) = t + f (0)2
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Therefore t = f (f (t)) = f (0) = t + f (0)2 , i
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f (0) = 0
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Consider now the two cases:
First case f (1) = 1
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Clearly
in this case we have f (y) = y for every real y
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Plugging x = −1 gives f (−1 + f (y)) = 1 + y, and after taking squares
(1 + y)2 = f (−1 + f (y))2 = (−1 + f (y))2 = 1 − 2f (y) + f (y)2 = 1 − 2f (y) + y 2
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It is easy to verify that f (x) = x and f (x) = −x are indeed the solutions
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(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
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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
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Plugging
in y = −1 we get f (x) = −f (−x)
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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) + 2f (v) + f (1) + f (2uv + u)
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e
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(1)

Plugging in v = −1/2 we get 0 = 2f (−u/2) + f (u) = −2f (u/2) + f (u)
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Now (1) reduces to f (2uv + u) = f (2uv) + f (u)
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Since f (0) = 0, it trivially holds that f (x + y) = f (x) + f (y) when one of x and y is 0
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Does there exist a function f : R → R such that f (f (x)) = x2 − 2 for every real
number x?
Solution
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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
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Assume
the contrary
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Assume that
g(c) = y
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If y = a then from a = g(a) = g(y) = c we get a contradiction
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Thus g(c) = d and g(d) = c
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Let x0 ∈ {a, b}
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Similarly if x1 ∈ {a, b, c, d} we get f (x1 ) ∈ {a, b, c, d}, and now we will
prove that this is not possible
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Then f (a) = f (f (c)) = g(c) = d which is
clearly impossible
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However we then have f (d) = f (f (c)) = g(c) = d, which is a contradiction, again
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△
Problem 8
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Solution
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The idea is to find y such that
x
≥ 0 we can find
yf (x) = x + y and use this to determine f (x)
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However this is a contradiction since we got
that f (x) > 1 implies f (x) = 1
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Assume that f (x) < 1
for some x
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Let us prove that f is decreasing
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Assume that f (x) = 1 for every x ∈ (0, a) (a > 0)
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This means that the function is decreasing and hence it
is injective
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Notice that
x + y > yf (x), therefore
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)) ,
i
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f (x) = f f yf (x) (x + y − yf (x))
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If we plug f (x) = a we get

f (y) =

1
,
1 + αz

1 − f (a)
, and according to our assumption α > 0
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△
1 + αx

where α =

Problem 9
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6

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imocompendium
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Let us first solve the problem under the assumption that g(α) = 0 for some α
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Then the given
equation becomes f (x + g(y)) = (α + 1 − y)f (x) + (f (y) − f (α))x, so setting y = α + 1 we get
f (x + n) = mx, where n = g(α + 1) and m = f (α + 1) − f (α)
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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}
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If f (0) = 0 then putting y = 0 in the given
equation we obtain f (x + g(0)) = g(x), so we can take α = −g(0)
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By replacing x by g(x) in the given equation we obtain
f (g(x) + g(y)) = g(x)f (y) − yf (g(x)) + g(g(x)) and, analogously, f (g(x) + g(y)) = g(y)f (x) −
xf (g(y)) + g(g(y))
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In particular, g is injective and f is surjective, so there exists c ∈ R such that f (c) = 0
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(1)

Plugging y = c in (1) we get g(g(x)) = g(c)f (x) − ax + g(g(c)) + ac = kf (x) − ax + d
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For y = 0 we have g(x)b + kf (x) = af (x) + kb,
whence
a−k
g(x) =
f (x) + k
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From the surjectivity of f it follows that g is
surjective as well, so it takes the value 0
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(IMO 1992, shortlist) Find all functions f : R+ → R+ which satisfy
f (f (x)) + af (x) = b(a + b)x
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This is a typical example of a problem that is solved using recurrent equations
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It follows from the
given equation in f that xn+2 = −axn+1 + b(a + b)xn
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In order to have xn ≥ 0 for all
n we must have λ2 = 0
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Since x0 was arbitrary, we
conclude that f (x) = bx is the only possible solution of the functional equation
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△
Problem 11
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Find the largest real number α
such that for all functions f ∈ F : f (x) ≥ α · x
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We clearly have that ∈ F , hence α ≤
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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
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2
2
2
Let us constract a recurrent relation for αk
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From the
given inequality we have
f (3x) ≥ f (f (2x)) + x ≥ αk f (2x) + x ≥ αk · αk · 2x + x = αk+1 · 3x
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Let us prove that limn→+∞ αn =
...

3
2
1
It is easy to prove that the sequence αk is increasing and bounded above by
...
e
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△
3
2
This means that αn+1 =

Problem 12
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Solution
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First
taking y = x and substituting g(0) = a we get f (2x) = 4h(x) − a
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Now the original equation can be written
2
as
2 h

x−y
x+y
+h
2
2

+ h(x − y) + b = h(x) + h(y)
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These ”longer” linear expressions can be easily handled if we express
functions in form of the sum of an even and odd function, i
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H(x) = He (x) + Ho (x)
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(3)

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)
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Mathematical induction yields He (r) = αr2 , for every rational
number r
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Similar method gives the simple relation for
Ho
Ho (x + y) + Ho (x − y) = 2Ho (x)
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Thus h(x) = αx2 + βx + b and substituting for f
and g we get:
f (x) = αx2 + 2βx + 4b − a, g(x) = αx2 + a
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Problem 13
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Solve the same problem for the case f : R → R
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It is not hard to see that for x = y = 0 we get (f (0)−1)2 = 0, i
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f (0) = 1
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We will
separate this into two cases:
1◦ Let f (−1) = 0
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e
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8

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imocompendium
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Namely the expression on the left-hand side is symmetric, while the one on the right-hand side
is not
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(4)

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)
...
e
...
This means that we have
two cases here as well:
1
...
Setting
−y instead of y in the initial equality gives f (xy) = f (x)f (y) − f (x − y) + 1, hence
f (x + y) = f (x − y), for every two rational numbers x and y
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However this is a contradiction with f (1) = 0
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1
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It is clear
that we should do the substitution g(x) = 1 − f (x) because the previous equality gives
g(−x) = −g(x), i
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g is odd
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(6)

Setting −y instead of y we get −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)
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This is a the Cauchy equation and since
the domain is Q we get g(x) = rx for some rational number r
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2◦ Let f (1) = 1
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This means that f (x) ≡ 1 and
this function satisfies the given equation
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Notice that we haven’t used that the range is Q,
hence we conclude that for all rational numbers q f (q) = q + 1, or f (q) ≡ 1
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Assume that f (q) ≡ 1
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Substitute x = y in
(6) and use g(2x) = 2g(x) to get g(x2 ) = −g(x)2
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Hence if y > x, i
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y = x+ r2 we have g(y) = g(x)+ g(r2 ) ≤ g(x), and the function
is decreasing
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It
is easy to verify that so obtained functions satisfy the given functional equation
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(IMO 2003, shortlist) Let R+ denote the set of positive real numbers
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Marko Radovanovi´ : Functional Equations
c

9

Solution
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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
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One of the properties of the solution suggested above is f (x) =
t
f (1/x), and proving this equality will be our next step
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(7)

In particular, for s = 1 the last equality yields f (t) = f (1/t); hence f (t) ≥ f (1) = 2 for each
1
t
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Now it follows by induction
from (7) that g(tn ) = g(t)n for every integer n, and therefore g(tq ) = g(t)q for every rational q
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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 (tr ) = ar + a−r for
every real r
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△

Problem 15
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Solution
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Let us prove that this is the only
solution
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e
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With this we have found the upper bound for f (x)
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Similarly we get
√
f (x)2 = xf (x + 1) + 1 ≤ x( 2(x + 1)) + 1 < 21/4 (1 + x)2
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However this is shown in the same way as the previous two inequalities
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This implies f (x) ≤ x + 1 for
every real number x ≥ 1
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We will use the
f (x)2 − 1
= f (x + 1) ≥ 1, i
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similar argument
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We further have f (x) = 1 + xf (x + 1) > 1 + x x + 2 > x3/2 and
similarly by induction
k
f (x) > x1−1/2
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Now again from the given equality we get f (x)2 =
1
1 + xf (x + 1) ≥ (x + 1/2)2 , i
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Using the induction we get f (x) ≥ x + 1 − k ,
2
and passing to the limit we get the required inequality f (x) ≥ x + 1
...
(IMO 1999, probelm 6) Find all functions f : R → R such that
f (x − f (y)) = f (f (y)) + xf (y) + f (x) − 1
...
imo
...
yu, www
...
com

10

Solution
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e
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We will determine the value of the function on
A
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From the given equality we have f (0) = f (x) + x2 + f (x) − 1,
i
...

c + 1 x2
− ,
f (x) =
2
2
where f (0) = c
...
Setting x = y = 0 in the
original equation we get f (−c) = f (c) + c − 1, hence c = 0
...
Since the range of the function (on x)
on the right-hand side is entire R, we get {f (x − c) − f (x) | x ∈ R} = R, i
...
A − A = R
...
Now we have
f (x) =
=

f (y1 − y2 ) = f (y1 − f (z)) = f (f (z)) + y1 f (z) + f (y1 ) − 1
x2
f (y1 ) + f (y2 ) + y1 y2 − 1 = c −
...
It is easy to show that the function f (x) = 1 −
2
satisfies the given equation
...
Given an integer n, let f : R → R be a continuous function satisfying f (0) = 0,
f (1) = 1, and f (n) (x) = x, for every x ∈ [0, 1]
...

Solution
...
The idea for
what follows is clear once we look at the graphical representation
...
Let us prove that formally
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The continuity on
[0, x1 ] implies that there is some c such that f (c) = f (x2 ), which contradicts the injectivity of f
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x < f (n)(x) = x
...
Hence for each x ∈ [0, 1] we must have f (x) = x
...
Find all functions f : (0, +∞) → (0, +∞) that satisfy f (f (x) + y) = xf (1 + xy)
for all x, y ∈ (0, +∞)
...
Let us prove that the function
x
is non-increasing
...
We will
yf (y) − xf (x)
consider the expression of the form z =
since it is positive and bigger then f (y)
...
Hence the function is non-decreasing
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Let f (1) = 1
...
Therefore the function is periodic on the interval (1, +∞),
and since it is monotone it is constant
...
Thus we must have f (1) = 1
...

that f (x) = for x > 1
...
If f (x) <
we have
If f (x) >
x
x
x
1
1
1
f f (x) − + 1 ≥ f (1) = 1, and xf (x) < 1
...
If x < 1, plugging y = we get
x
x
x
Solution
...
e
...
This means that f (x) =
x
x
x
x
x
for all positive real numbers x
...
(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 sufficiently large m this implies f (x + m) < 0
...
Assume that such function exists
...
Also from the given
get f (x) − f (x + y) ≥
f (x)
equality we can conclude that
f (x) − f (x + y) ≥

f (x)y

...
Notice that for
0 ≤ k ≤ n − 1 the following inequality holds
k 1
f x+ n n
k
1
k+1
f x+
≥
−f x+
≥
,
1
k
n
n
2n
f x+ n + n

1
and adding similar realitions for all described k yields f (x)−f (x+1) ≥ which is a contradiction
...
Let f : N → N be a function satisfying
f (1) = 2,

f (2) = 1,

f (3n) = 3f (n),

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

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

Solution
...
For this situation the base 3 is doing the job
...
Clearly the given equation can have only one solution
...

Now we see that f (n) is obtained from n by changing each digit 2 by 1, 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)
...
The answer is 127
...
(BMO 2003, shortlist) Find all possible values for f
2003
the function satisfying the conditions:
(i) f (xy) = f (x)f (y) za sve x, y ∈ Q;
(ii) f (x) ≤ 1 ⇒ f (x + 1) ≤ 1 za sve x ∈ Q;
(iii) f

2003
= 2
...
imo
...
yu, www
...
com

Solution
...
Now (i)
implies that for x = y = 1 we get f (1) = 0 and similarly for x = y = −1 we get f (−1) = 1
...
For f (x) ≤ f (y) from f
f (y) = f (x) we have that
x
y
y
f
≤ 1, and according to (ii) f
+ 1 ≤ 1
...
Now you might wonder how did we get
this idea
...
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)
...
There are a and b such that ap + bq = 1 implying f (1) = f (ap + bq) ≤ max{f (ap), f (bq)}
...
From (iii)
we have
f (2003)
2003
=
= 2,
f
2002
f (2)f (7)f (11)f (13)
hence only one of the numbers f (2), f (7), f (11), f (13) is equal to 1/2
...

=
2003
f (2003)
2003
1
If f (2) = 1/2 then f
= , otherwise it is 1
...
For the first value it is the
multiplicative function taking the value 1/2 at the point 2, 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
...
△
Problem 22
...
Find all f : G → I such that for all x, y, z ∈ I
the following statements hold:
(i) f (f (x, y), z) = f (x, f (y, z));
(ii) f (x, 1) = x, f (x, y) = f (y, x);
(ii) f (zx, zy) = z k f (x, y) for every x, y, z ∈ I, where k is a fixed real number
...
The function of several variables appears in this problem
...
From the condition (ii) we get f (1, 0) =
f (0, 1) = 0, and from (iii) we get f (0, x) = f (x, 0) = xk f (1, 0) = 0
...
Assume therefore that 0 < x ≤ y < 1
...
This implies
f (x, y) = f (y, x) = y k f 1,

x
= y k−1 x
...
Let us find all
1
possible values for k
...
From the condition (i), and the already obtained
2

Marko Radovanovi´ : Functional Equations
c

13

results we get
f f x,

1
1
,y = f x
2
2

k−1

, y = f x, f

1
2

1
= f x, y k−1
...
e
...
For
2
k = 1 the solution is f (x, y) = min(x, y), and for k = 2 the solution is f (x, y) = xy
...
△
Problem 23
...

Solution
...
Another
useful idea appears in this problem
...
Our goal is to prove that Sd = R
...
Let us prove
that for x ∈ Sd we have x + d ∈ Sd as well
...
e
...
Let us prove that the sets Sd′ are empty, where d′ < d
...
e
...
Let us use this to get the contradiction
...
Assume the contrary
...
By further induction
we prove that every y satisfying
x + k(d − d′ ) ≤ y < x + (k + 1)(d − d′ ),
can’t be a member of Sd′
...
Similarly if d′ > d switching the roles of d and d′ gives a contradiction
...
△
Problem 24
...

Solution
...
We have to use another approach to this problem
...

Setting n = 1 gives h(h(1)) + h(2) = 3, therefore h(h(1)) ≤ 2 and h(2) ≤ 2
...
Then h(h(1)) = 2
...
Let h(1) = k
...

This means that k = 3, hence h(3) = 1
...
This means that there are no solutions in this case
...
Then h(h(1)) = 1
...
Setting
n = 3, 4, 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
...
Clearly there is at most one function
satisfying the given equality
...
The solution is
h(n) = ⌊nα⌋ + 1,
√
−1 + 5
(this constant can be easily found α2 + α = 1)
...
△

14

Olympiad Training Materials, www
...
org
...
imocompendium
...
(IMO 2004, shortlist) Find all functions f : R → R satisfying the equality
f (x2 + y 2 + 2f (xy)) = f (x + y)2
...
Let us make the substitution z = x + y, t = xy
...
Define g(x) = 2(f (x) − x)
...


(8)

Let us set c = g(0) = 2f (0)
...


(9)

If c < 0, then taking z such that z 2 + c = 0, we obtain from (9) that f (z)2 = c/2, which is
impossible; hence c ≥ 0
...


(10)

If g is a constant function, we easily find that c = 0 and therefore f (x) = x, which is indeed a
solution
...
For some
sufficiently large K and each u, v ≥ K with v 2 − u2 = d the equality u2 + g(a) = v 2 + g(b) by
(8) and (10) implies f (u) = f (v)
...

u
Therefore every value from some suitably chosen segment [δ, 2δ] can be expressed as g(u) − g(v),
with u and v bounded from above by some M
...
By the above considerations,
there exist u, v ≤ M such that g(u) − g(v) = y 2 − x2 , i
...
, x2 + g(u) = y 2 + g(v)
...
Moreover, if we assume √
w
...
o
...
that 4M ≥ c2 , we
conclude from (10) that f (x) = f (y)
...
Setting x > N in (9)
we obtain k 2 = k, so k = 0 or k = 1
...
Hence g(u) = 2f (u) −
2u ≥ −2 − 2u for u ≤ −N , which implies that g is unbounded
...
Therefore f (z) = ±k for
each z
...
Assume k = 1
...
Suppose that f (t) = −1
for some t < 2
...
If also t − g(t) ≥ 0, then for some z ∈ R we have
z 2 = t − g(t) > 4t, which by (8) leads to f (z)2 = f (z 2 + g(t)) = f (t) = −1, which is impossible
...
On the other hand, 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
...
△

4

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
...


Marko Radovanovi´ : Functional Equations
c

15

1
...

2
...

3
...

4
...

5
...
Find all monotone functions f : R → R such that
f (x + f (y)) = f (x) + y n
...
(USA 2002) Find all functions f : R → R which satisfy the equality f (x2 − y 2 ) = xf (x) −
yf (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
...
Find all continuous functions f : R → R such that f (xy) = xf (y) + yf (x)
...
(IMO 1983, problem 1) Find all functions f : R → R such that
(i) f (xf (y)) = yf (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
...

11
...
Determine f

1

...
(IMO 1996, shortlist) Let f : R → R be the function such that |f (x)| ≤ 1 and
f x+

13
1
1
+ f (x) = f x +
+f x+

...

13
...


14
...

y

Olympiad Training Materials, www
...
org
...
imocompendium
...
(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)))
...

17
...

18
...
Find all
functions f : S → S that satisfy the following two conditions:
(i) f (x + f (y) + xf (y)) = y + f (x) + yf (x) for all x, y ∈ S;
f (x)
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) = y α f (x/2) +
xβ f (y/2), za sve 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)
...
(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)) = mf (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?
24
...
Assume that the function f : N → N satisfies f (n + 1) > f (f (n)), for every n ∈ N
...

26
...

27
...

28
...

29
...

30
...

31
...
Prove that f (1996x) = 1996f (x) for every x ∈ R
...
Find all functions f : R → R that satisfy:
(i) f (x + y) = f (x) + f (y) for every two real numbers x and y;
(ii) f

f (x)
1
= 2 for x = 0
...
(IMO 1989, shortlist) A function f : Q → R satisfy the following conditions:
(i) f (0) = 0, f (α) > 0 za α = 0;
(ii) f (αβ) = f (α)f (β) i f (α + β) ≤ f (α) + f (β), za sve α, β ∈ Q;

(iii) f (m) ≤ 1989 za m ∈ Z
...

34
...

35
...


36
...

37
...

38
...


Olympiad Training Materials, www
...
org
...
imocompendium
...
(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;

(iii) If x = 0 then f x +

1
x2

= f (x) + f

1
x

2

?

40
...

41
...

f

42
...

43
...

44
...

45
...


46
...

47
...
Find all natural numbers n ≤ 1998 such that f (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
...


Marko Radovanovi´ : Functional Equations
c

19

49
...
Prove that f (x, x) = 1, f (x, −x) = 1, and f (x, y)f (y, x) =
1
...
Find all functions f : N × N → R that satisfy
f (x, x) = x,

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

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

---