Functional Equations Worksheet
1. Find all functions 𝑓 which satisfy the following:
a. 𝑓(𝑥)𝑓(𝑦) = 𝑓(𝑥) + 𝑓(𝑦) for all real 𝑥.
𝑥
b. 𝑓 (𝑦) = 𝑓(𝑥)𝑓(𝑦) for all real non-zero values.
c. 𝑓(2𝑥) = 𝑓(𝑥) for all positive integers.
d. 𝑓(1) = 2 and 𝑓(𝑥𝑦) = 𝑓(𝑥)𝑓(𝑦) − 𝑓(𝑥 + 𝑦) + 1 for all 𝑥, 𝑦 ∈ ℤ (i.e. over all
integers).
e. 𝑓(𝑥) = 𝑓(𝑦) for all real 𝑥, 𝑦.
2. Find all functions 𝑓 such that 𝑓(𝑥) + 𝑥 𝑓(1 − 𝑥) = 𝑥 2 for all real 𝑥.
3. [Based on a BMO1 Question 1992] Let 𝑓 be a function mapping positive integers to positive
integers. Suppose that:
𝑓(𝑛 + 1) > 𝑓(𝑛) and 𝑓(𝑓(𝑛)) = 3𝑛 for all positive integers 𝑛.
Our ultimate goal is to determine 𝑓(1992).
a.
b.
c.
d.
e.
f.
What is 𝑓(𝑓(1))?
Explain why 𝑓 cannot be the identity function, and hence why 𝑓(𝑛) > 𝑛.
Use this to show that 𝑓(1) = 2.
Work out 𝑓(2), 𝑓(3) and 𝑓(6). What must 𝑓(4) and 𝑓(5) therefore be?
Construct a table finding the values for 𝑓(1) through to 𝑓(30). What patterns can
you see? (Hint: look down your table from each power of 3, i.e. 𝑓(1), 𝑓(3), 𝑓(9), …)
Hence determine 𝑓(1992).
4. [Source: BMO1] 𝑓 is a function from ℤ+ to ℤ+ , where ℤ+ is the set of non-negative integers,
which has the following properties:
a. 𝑓(𝑛 + 1) > 𝑓(𝑛) for each 𝑛 ∈ ℤ+ ,
b. 𝑓(𝑛 + 𝑓(𝑚)) = 𝑓(𝑛) + 𝑚 + 1 for all 𝑚, 𝑛 ∈ ℤ+ .
Find all possible values of 𝑓(2001).
[Hint: The key here is to try and establish that 𝑓(0) = 1. Once you’ve managed to do that,
the rest of the problem becomes relatively easy]
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Functional Equations Worksheet - ANSWERS
1. Find all functions 𝒇 which satisfy the following:
a. 𝒇(𝒙)𝒇(𝒚) = 𝒇(𝒙) + 𝒇(𝒚) for all real 𝒙 and 𝒚.
Let 𝑥 = 𝑦. Then 𝑓(𝑥)2 = 2𝑓(𝑥). Then 𝑓(𝑥)(𝑓(𝑥) − 2) = 0. So either 𝑓(𝑥) = 0 or
𝑓(𝑥) = 2. Both of these satisfy the equation.
𝒙
b. 𝒇 (𝒚) = 𝒇(𝒙)𝒇(𝒚) for all real non-zero values.
Let 𝑥 = 𝑦. Then 𝑓(1) = 𝑓(𝑥)2 . Let 𝑥 = 1. Then 𝑓(1) = 𝑓(1)2 . So 𝑓(1)(𝑓(1) − 1) =
0. Thus 𝑓(1) = 0 or 𝑓(1) = 1.
Before this second substitution, we had that 𝑓(𝑥) = ±√𝑓(1) for all 𝑥 ≠ 0.
In the case where 𝑓(1) = 0, then 𝑓(𝑥) = 0.
In the case where 𝑓(1) = 1, then 𝑓(𝑥) = ±1. But 𝑓(𝑥) = −1 is not valid for all 𝑥,
because we know 𝑓(1) ≠ −1.
Thus 𝑓(𝑥) = 0 or 𝑓(𝑥) = 1 are solutions.
c. 𝑓(𝟐𝒙) = 𝒇(𝒙) for all positive integers.
Suppose 𝑓(1) = 𝑘1 . Then clearly 𝑓(2) = 𝑓(4) = 𝑓(8) = ⋯ = 𝑘1 . Similarly, if
𝑓(3) = 𝑘3 , then 𝑓(6) = 𝑓(12) = 𝑓(24) = ⋯ = 𝑘3.
Every positive integer can be expressed uniquely as the product of an odd number
and a power of 2, i.e. in the form 𝑐 ⋅ 2𝑑 where 𝑐 is odd and 𝑑 is an integer such that
𝑑 ≥ 0. Thus for each odd number we can associate a constant, say 𝑘𝑐 , such that
𝑓(𝑐 ⋅ 2𝑑 ) = 𝑘𝑐 for all integers 𝑑 ≥ 0. Since each integer can be expressed uniquely
in this form, each input maps to a single output (thus satisfying the requirements of
a function).
d. 𝒇(𝟏) = 𝟐 and 𝒇(𝒙𝒚) = 𝒇(𝒙)𝒇(𝒚) − 𝒇(𝒙 + 𝒚) + 𝟏 for all 𝒙, 𝒚 ∈ ℤ (i.e. over all
integers).
Letting 𝑦 = 1, 𝑓(𝑥) = 𝑓(𝑥)𝑓(1) − 𝑓(𝑥 + 1) + 1. This gives 𝑓(𝑥 + 1) = 𝑓(𝑥) + 1,
i.e. when the input increases by 1, the output increases by 1. Since 𝑓(1) = 2, then
inductively, 𝑓(𝑥) = 𝑥 + 1 for all natural numbers, and using 𝑓(𝑥) = 𝑓(𝑥 + 1) − 1,
we can go in the negative direction to show that the 𝑓(𝑥) = 𝑥 + 1 is also true of
𝑓(0) and negative 𝑥.
e. 𝒇(𝒙) = 𝒇(𝒚)
𝑓(𝑥) = 𝑐. If it wasn’t a constant function, then we could always say change 𝑦 to find
a value of 𝑓(𝑦) that is different to the LHS of the equation.
2. Find all functions 𝒇 such that 𝒇(𝒙) + 𝒙 𝒇(𝟏 − 𝒙) = 𝒙𝟐 for all real 𝒙.
As per the lecture slides, 1 − 𝑥 is a self-inverse function.
So letting 𝑦 = 𝑥 first:
𝑓(𝑦) + 𝑦 𝑓(1 − 𝑦) = 𝑦 2
(1)
and letting 𝑦 = 1 − 𝑥:
𝑓(1 − 𝑦) + (1 − 𝑦)𝑓(𝑦) = (1 − 𝑦)2 (2)
We can eliminate the 𝑓(1 − 𝑦) by doing (1) − 𝑦(2), giving:
𝑓(𝑦)(1 − 𝑦(1 − 𝑦)) = 𝑦 2 − 𝑦(1 − 𝑦)2
Simplifying, we obtain:
3𝑦 2 − 𝑦 3 − 𝑦
𝑓(𝑦) = 2
𝑦 +1−𝑦
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3. Let 𝒇 be a function mapping positive integers to positive integers. Suppose that:
𝒇(𝒏 + 𝟏) > 𝒇(𝒏) and 𝒇(𝒇(𝒏)) = 𝟑𝒏 for all positive integers 𝒏.
Our ultimate goal is to determine 𝒇(𝟏𝟗𝟗𝟐).
a. What is 𝒇(𝒇(𝟏))?
𝑓(𝑓(1)) = 3 by the definition given.
b. Explain why 𝒇 cannot be the identity function, and hence why 𝒇(𝒏) > 𝒏.
If it was the identity function, then 𝑓(1) = 1, and 𝑓(3) = 3. But we know
𝑓(𝑓(1)) = 3, but then 𝑓(1) = 3, a contradiction. All integers must be positive, and
by 𝑓(𝑛 + 1) > 𝑓(𝑛), our function must be ascending. But since 𝑓(1) ≠ 1, it must be
the case that 𝑓(1) > 1, and thus in the case where the function increases as slowly
as possible (adding 1 each time), 𝑓(𝑛) > 𝑛.
c. Use this to show that 𝒇(𝟏) = 𝟐.
If 𝑓(1) = 3, then 𝑓(𝑓(1)) = 𝑓(3) = 3, a contradiction since 𝑓(𝑛) > 𝑛.
If 𝑓(1) > 3, then 𝑓(𝑓(1)) > 𝑓(3) > 3. But we know that 𝑓(𝑓(1)) = 3, which is a
contradiction. Since 𝑓(1) > 1, then 𝑓(1) must be 2.
d. Work out 𝒇(𝟐), 𝒇(𝟑) and 𝒇(𝟔). What must 𝒇(𝟒) and 𝒇(𝟓) therefore be?
Using 𝑓(𝑓(𝑛)) = 3𝑛, then 𝑓(𝑓(1)) = 𝑓(2) = 3. 𝑓(𝑓(2)) = 𝑓(3) = 6.
𝑓(𝑓(3)) = 𝑓(6) = 9. Now since 𝑓(𝑛 + 1) > 𝑓(𝑛), the integers have to be
increasing each time. Since 𝑓(3) = 6 and 𝑓(6) = 9, then it must be that 𝑓(4) = 7
and 𝑓(5) = 8.
e. Construct a table finding the values for 𝒇(𝟏) through to 𝒇(𝟑𝟎). What patterns can
you see?
The quick way to generate the table is for the given 𝑛, trying to find the value in the
𝑓(𝑛) column, say 𝑓(𝑘). Then multiply that 𝑘 by 3 (it’s basically using the second
condition backwards).
𝑓(𝑛)
𝑛
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
2
3
6
7
8
9
12
15
18
19
20
21
22
23
24
𝑓(𝑛)
𝑛
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
25
26
27
30
33
36
39
42
45
48
51
54
55
56
57
The pattern is that after each power of 3, we have that same number of consecutive
numbers after it (e.g. when n=9, we have 9 consecutive numbers after it). Then
when we’re at double a power of 3, we start going up in 3s until the next power of 3
is reached.
We could show some of these properties by ‘induction’.
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For example we could show that 𝑓(3𝑛 ) = 2 ⋅ 3𝑛 (which we can see from the table):
In the base case, when 𝑛 = 0, we’ve already shown that 𝑓(30 ) = 𝑓(1) = 2. Now
assume that 𝑓(3𝑛 ) = 2 ⋅ 3𝑛 . Then 𝑓(𝑓(3𝑛 )) = 3 ⋅ 3𝑛 = 3𝑛+1 by definition of the
function, and by induction 𝑓(2 ⋅ 3𝑛 ) = 3𝑛+1. Applying the function to both sides of
the equation
𝑓(𝑓(2 ⋅ 3𝑛 )) = 𝑓(3𝑛+1 ). But again by definition of the function,
𝑓(𝑓(2 ⋅ 3𝑛 )) = 3 × 2 ⋅ 3𝑛 = 2 ⋅ 3𝑛+1 . Thus 𝑓(3𝑛+1 ) = 2 ⋅ 3𝑛+1 , and our proof is
complete.
f.
Hence determine 𝒇(𝟏𝟗𝟗𝟐).
The highest power of 3 below 1992 is 36 = 729. We know that 𝑓(729) = 2 ×
729 = 1458. Now now have 729 consecutive integers, so we have 𝑓(1458) =
1458 + 729 = 2187. At this point the values of 𝑓(𝑛) go up in 3s. Our value of 𝑛,
1992, is 1992 − 1458 = 534 away. So 𝒇(𝟏𝟗𝟗𝟐) = 𝟐𝟏𝟖𝟕 + (𝟑 × 𝟓𝟑𝟒) = 𝟑𝟕𝟖𝟗.
4. 𝒇 is a function from ℤ+ to ℤ+ , where ℤ+ is the set of non-negative integers, which has the
following properties:
a. 𝒇(𝒏 + 𝟏) > 𝒇(𝒏) for each 𝒏 ∈ ℤ+ ,
b. 𝒇(𝒏 + 𝒇(𝒎)) = 𝒇(𝒏) + 𝒎 + 𝟏 for all 𝒎, 𝒏 ∈ ℤ+ .
Find all possible values of 𝒇(𝟐𝟎𝟎𝟏).
Firstly, establish that 𝒇(𝟎) = 𝟏
Let 𝑓(0) = 𝑘, where 𝑘 is a non-negative integer (since the range of 𝑓 is ℤ+ ). When 𝑚 = 0
the condition (b) gives:
𝑓(𝑛 + 𝑘) = 𝑓(𝑛) + 1 (1)
If 𝑘 = 0 then 𝑓(𝑛) = 𝑓(𝑛) + 1, a contradiction. Hence 𝑘 ≠ 0.
From condition (a) we have that
𝑓(𝑛 + 𝑘 − 1) < 𝑓(𝑛 + 𝑘) = 𝑓(𝑛) + 1 (2)
However, if 𝑘 > 1 then 𝑛 + 𝑘 − 1 ≥ 𝑛 + 1 so that by (a)
𝑓(𝑛 + 𝑘 − 1) ≥ 𝑓(𝑛 + 1) ≥ 𝑓(𝑛) + 1 (3)
(2) and (3) are contradictory. Hence 𝑘 = 1 is the only possible solution.
Alternatively:
Substituting 𝑛 = 0 into (1) gives
𝑓(𝑘) = 𝑘 + 1
Substituting 𝑛 = 1 into (1) gives
𝑓(1 + 𝑘) = 𝑓(1) + 1
(4)
Substituting 𝑛 = 0 and 𝑚 = 𝑘 into condition (b) we get
𝑓(0 + 𝑓(𝑘)) = 𝑓(0) + 𝑘 + 1
i.e.
𝑓(𝑘 + 1) = 2𝑘 + 1 (5)
From (4) and (5), 𝑓(1) = 2𝑘. However if 𝑘 > 1, then condition (a) states that 𝑓(𝑘) > 𝑓(1)
so that 𝑘 + 1 > 2𝑘, a contradiction. Thus 𝑘 = 1.
Now use that 𝒇(𝟎) = 𝟏 to find 𝒇(𝟐𝟎𝟎𝟏)
We’re pretty much there. Substituting 𝑚 = 0 into (b) we get:
𝑓(𝑛 + 1) = 𝑓(𝑛) + 1
Hence 𝑓(2001) = 2002.
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