Download EGYPTIAN FRACTIONS – REPRESENTATIONS AS SUMS OF UNIT

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EGYPTIAN FRACTIONS – REPRESENTATIONS
AS SUMS OF UNIT FRACTIONS
Hans Humenberger
Faculty for Mathematics
University of Vienna
Oskar-Morgenstern-Platz 1
A – 1090 Vienna, Austria
[email protected]
1. INTRODUCTION
The following paper presents exercises built around the very old
mathematical topic, Egyptian Fractions. Teachers can use these exercises
and this topic (roughly: fractions built up as a sum of unit fractions) in the
classroom to stimulate learning, or in extracurriculr settings (e.g., a
mathematics club project).
The Egyptians did not have an elaborated denominational number
system; they had only symbols for their numbers (hieroglyphics). A bar | for 1,
two bars || for 2, for 10 they used ∩ and other symbols for the powers of 10
until 106. They put an oval (this was the hieroglyphic for “mouth”, in this
connection meaning “part”) above the number symbol to indicate the
corresponding reciprocal number (Figure 1).
Figure 1
These reciprocals of natural numbers are called unit fractions 1.
Using the oval they were able to write unit fractions and sums of unit
fractions by writing them side by side (Figure 2). One could also write
3 = 1 + 1 , both possibilities are shorter than 3 = 1 + 1 + 1 where
10 5 10
10 10 10 10
three symbols would be needed. Using this principle the Egyptians would
have had to write the oval above the symbol 14 times when they meant the
fraction 14/25, which is not very efficient, especially when the symbols had
to be carved in stone! Moreover, doing so requires a lot of space for the
1 For 1/2 and the non unit fraction 2/3 they had special symbols, often also for 1/3, 1/4 and
3/4. This is documented differently in the literature, but is not addressed here.
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fraction to be well arranged. But if one knows that 14 = 1 + 1 + 1 , one
25 2 17 850
only needs three symbols: 1/2, 1/17, 1/850. The Egyptians used this principle
and displayed their fractions as a sum of unit fractions. A table on the
famous Papyrus Rhind (approximately 1650 BCE, one of the oldest
conserved “pieces of mathematics”) which displays – for odd numbers n
from 5 to 101 – the fractions 2/n as a sum of different 2 unit fractions. This
topic must have been important in the mathematics of the old Egyptians
since it takes about one third of the Papyrus.
Figure 2
The study of Egyptian Fractions can be deepened and extended in
several ways, and there are many fascinating historical presentations in the
literature as well as on the internet, thus students can easily research other
aspects of this topic.
2. TASKS FOR INTRODUCTION
Interesting sums of unit fractions have to consist of different unit
fractions. So 3 = 1 + 1 + 1 would not be an interesting representation in the
5 5 5 5
manner of the Egyptians. But nevertheless such representations (see a
through g, below) can be used for students’ activities [6] and should be
appropriate for some students in high schools and colleges. But they may
need a hint or two.
a. Is it possible to represent any unit fraction as a sum of two unit
fractions?
1 1
(Yes: 1
n = 2n + 2 n )
b. Is the sum of two unit fractions (may be after reducing) always a unit
fraction?
(No: 1 + 1 = 5 )
2 3 6
2 Even if one claims that the unit fractions must be different such a representation needs not
to be unique as we see when we consider the different representations of 3/10 above.
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c. Is the product of two unit fractions always a unit fraction?
(Yes: 1a ⋅ 1 = 1 )
b a⋅b
d. Can every unit fraction be represented as a product of two unit fractions?
1 1 1
(Yes: = ⋅ ; if one wants to omit this trivial case:
n n 1
Yes, if n is not prime.)
e. Can every fraction be represented as a quotient of two unit fractions?
m 1 1
(Yes:
= ÷ )
n n m
f. Can every fraction be represented as a sum of two unit fractions?
(No: surely not numbers > 2, see section 3.3)
g. Can every fraction < 1 be represented as a sum of unit fractions?
m
1
(Yes:
is the sum of m summands )
n
n
The following questions (h, i, j) might be more appropriate for college
mathematics majors.
h. Is there for every fraction < 1 a representation as a sum of two unit
fractions?
(No: e.g. with 4/5 one will not find such a representation; how can we
formulate that fact so that it is substantial reasoning? – a task for
students, not really easy but also not asking too much: e.g. because of
1 1 2 4
1
+ = < at least one of the two unit fractions would have to be ;
3 3 3 5
2
4 1 3
but
= + . If students do not find this counterexample by
5 2 10
themselves the teacher could give this hint. Then the task for the students
is just to give reasons for the impossibility.)
i. Is there for every unit fraction a representation as a sum of two different
unit fractions?
1 1 1 1 1 1 1 1 1
(By trial and error one finds :
. . .
= + , = + ,
= +
2 3 6 3 4 12 4 5 20
1
1
1
and in general:
, this can be confirmed by
=
+
n n + 1 n ⋅ (n + 1)
“calculating”. Here students must be creative to discover this structure,
1
but it is not very hard to find because
is the biggest unit fraction
n +1
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that can be “extracted” from
fraction (!)
1
, and as the “remainder” one gets the unit
n
1
.)
n ⋅ (n + 1)
j. Is there for every unit fraction a representation as a sum of two, three,
four or more different unit fractions?
(If one uses the results of i when having a look at the unit fractions
1
and
n +1
1
respectively, it is clear that every unit fraction can be
n ⋅ (n + 1)
represented as a sum of two, three, four or more different unit fractions.)
Although students may not find general answers to questions h, i,
and j they may enjoy and learn from the effort. But some hints may be
needed. Heinrich Winter, a famous German mathematics educator
established the term “discovering with practicing and practicing with
discovering”.
3. ADDITIONAL QUESTIONS
The following two questions arise quite naturally and have played a
role in history. The corresponding answers and insights can be achieved by
elementary means but considerable support may be needed from the teacher.
1. Can every fraction between 0 and 1 be represented as a sum of different
unit fractions? Is such a representation unique (not respecting the order
of the unit fractions)?
2. Are there always representations as a sum of two different unit fractions
when thinking of positive fractions with numerator 2 and 3 respectively
2
3
(i.e. and )?
n
n
Next we deal with question 1.
3.1. FRACTIONS BETWEEN 0 AND 1
unique
for
2
:
9
One can see immediately that such a representation needs not to be
3
with an example. Above we had different representations for
or
10
1 1 2 1 1
+ = = + .
6 18 9 5 45
When working on such examples (finding different representations
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as a sum of different unit fractions) one can get several hints that probably
every fraction between 0 and 1 has a representation as a sum of different unit
fractions. The teacher has to organise the learning process so that students
can discover many “things” by themselves. A simple method is the
following:
One looks for the biggest unit fraction that is part of the initial
fraction or part of the current remainder respectively, e.g. for the fraction
13
13
1
: The biggest unit fraction 3 that one can “extract” out of
is ,
17
17
2
13 1 9
calculating the remainder yields
− = , therefore we have
17 2 34
13 1 9
9
. The biggest unit fraction in the remainder
is
= +
17 2 34
34
9
9 1
9 1 1
; we
=
= ; calculating again the remainder yields
− =
34 + 2 36 4
34 4 68
13 1 1 1
arrived at a unit fraction, thus we have:
= + + . This procedure of
17 2 4 68
“extracting” the biggest possible unit fraction has several important aspects:
1. Students can discover it by themselves.
2. It provides many opportunities for calculating and exercising.
3. It gives the crucial hint to the general case and proof (see below) of the
fact that every fraction 0 < z / n < 1 has a representation as a sum of
different unit fractions. We have here an example of a so called
substantial learning environment [7].
Many representations of fractions as sums of unit fractions are built
according to this principle: Using the first unit fraction we have the one
1
13
which lies next to the initial fraction, e.g.,
lies next to
. If one would
2
17
13
by one unit fraction as good as possible, one
be forced to approximate
17
1
would take
. Th best unit fraction approximation for the remainder
2
13 1
1
( − ) is , and 1 is the best unit fraction for 13 − 1 − 1 .
17 2 4
68
17 2
4
3Here one can find it without calculating. In general one can find the biggest unit fraction in
z/n by increasing n to the next multiple of z:
13
17
→
13
26
=
1
. Here one had to increase n = 17
2
by k = 9 to n + k = 26, see below. This way will help us in the proof below.
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Theorem 1: Every fraction 0 <
z
< 1 has a representation as a sum of
n
different unit fractions.
To prove this theorem we show on the one hand that the numerators
of the reduced remainder fractions get smaller in every step (this implies that
the procedure of “extracting” the biggest possible unit fraction ends up at
numerator 1 and therefore we have a sum of unit fractions), and on the other
hand that the biggest extracted unit fractions are always bigger than the
remainders (this implies that the unit fractions in the sum are all different
because the remainder fraction is too small for extracting the same unit
fraction a second time).
z
From the fraction 0 < < 1 , (i.e. z < n ) we subtract the biggest
n
z
1
possible unit fraction . The remainder then is 1 (reduced) and we have
n1
m
z 1 z1
= + . This biggest possible unit fraction one can find by increasing
n m n1
n by k < z to the next multiple of z (after reducing there should be 1 in the
1
z
n+k
numerator):
, that means m =
and k = z ⋅ m − n
=
m n
+k
z
z|
respectively. This yields
z1
z 1 z⋅m − n
k
= − =
=
n1 n m
nm
nm
(1)
Remark: If students find the biggest unit fraction by increasing the
denominator by k and then calculate the remainders it is possible that they
discover by themselves that in the numerator of the remainder fraction
there is always the increase number k (maybe the fraction can be reduced).
9
13 1 9
For instance above we had
as remainder fraction
− = , and the
34
17 2 34
increase number was k = 9. If students have discovered this then (1) is just
a general confirmation “ex post”.
(1) is a crucial relation. On the one hand because of k < z and z < n ,
altogether k < n , it immediately implies
z1 1
< .
n1 m
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On the other hand, because
z1
is reduced, this implies z1 ≤ k . Since
n1
k < z , we get
(3)
z1 < z .
1
is
m
too big for extracting it another time”) and (3) implies that (after at most
z − 1 steps, the numerator gets less in every step by at least 1) one ends up
at numerator 1, that means we have a unit fraction as remainder and
therefore altogether a representation as a sum of different unit fractions.
There are also more complicated proofs for Theorem 1 but this one 4
can be grasped by many students.
This algorithm to find a representation as a sum of different unit
fractions is called “the Fibonacci algorithm” which can be found in the book
“Liber Abaci” by Leonardo of Pisa (approx. 1180 – approx. 1240; he is
called Fibonacci: “son of Bonaccio”). It is a “greedy” algorithm in the sense
of always extracting the biggest possible unit fraction. There are also other
algorithms for getting a representation as a sum of unit fractions ([3], [4], [5]
or [1, p. 421 ff]). Such algorithms may be useful because the Fibonacci
algorithm is not “optimal” in some sense: Sometimes it would be possible to
have a representation with a smaller number of unit fractions, sometimes
there would be a representation with a smaller maximum denominator etc.
But here we want to concentrate on the Fibonacci algorithm, the correctness
of which was proved first by J. J. Sylvester and therefore is often also called
Fibonacci-Sylvester algorithm (cf. [4, p. 206]). Now we come to question 2.
(2) implies that all the unit fractions are different (“the unit fraction
3.2. FRACTIONS OF THE FORM 2
n
We know already from i) that all unit fractions can be written as a
sum of two different unit fractions. When considering fractions of the form
2
one can see that these fractions can be written as a sum of different unit
n
fractions.
2 can be written as a sum of
Theorem 2: Every fraction of the form n
two different unit fractions.
We prove this theorem by considering two different cases:
4 I thank E. Wittmann (Dortmund) for giving me a hint for simplifying.
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2 1
= is a unit fraction and from i we know that there is
n k
such a representation.
2 1 1 2 1 1
= + ,
= + ,
• If n = 2k − 1 is odd one can find examples
3 2 6 5 3 15
2 1 1 2 1 1
2
1
1
; this
= + , = + , . . . and in general
= +
7 4 28 9 5 45
2k − 1 k k (2k − 1)
relation was probably already known by the ancient Egyptians (the
examples of the Papyrus Rhind are of this type). It can be found because
1
2
is the biggest unit fraction that can be extracted from
and as a
k
2k − 1
1
remainder fraction one gets the unit fraction
.
k (2k − 1)
• If n = 2k is even
The proof is an application of the mathematical principle called
distinguishing different cases. It provides a systematical means of
considering all possible cases – here demonstrated by an elementary
example; we will face this principle several more times in this paper.
3.3 FRACTIONS OF THE FORM 3
n
We can also consider fractions with numerator 3 because one can
come to a general survey with elementary methods. This is not the case
when we think of bigger numerators. There we quickly come to (up to now)
unsolved problems in number theory; this means that the case “numerator 3”
is near to interesting ideas in mathematical research.
3
Let us think of fractions of the form . Which of them can be
n
represented as a sum of two different unit fractions? – Let us again
distinguish the corresponding cases:
3 1
• For n = 3k one gets unit fractions = , according to i these have such
n k
a representation.
1
• For n = 3k − 1 one can extract the biggest possible unit fraction , the
k
3
1
1
result of it is again a unit fraction:
.
= +
3k − 1 k k (3k − 1)
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•
The case n = 3k + 1 is not as easy because one cannot extract
one can extract the next smaller unit fraction
1
. But
k
1
, and with
k +1
3
1
2
and the knowledge from above it is clear
=
+
3k + 1 k + 1 (k + 1)(3k + 1)
3
can be written as a sum of at most three different unit
that
3k + 1
2
fractions because
surely can be written as a sum of two
(k + 1)(3k + 1)
different unit fractions. We can say even more: If k = 2l − 1 is odd we
2
2
1
, and we have a
have
=
=
(k + 1)(3k + 1) 2l (3k + 1) l (3k + 1)
3
representation of
as a sum of two different unit fractions.
3k + 1
Considering numerator 3, i.e. fractions of the form
3
, we summerize for
n
n > 1:
• All these fractions can be written as a sum of at most three different unit
fractions.
• Only in the case n = 3k + 1 and k = 2l is even (that means n = 6l + 1 )
we do not know whether the maximum number 3 of unit fractions is
really needed. Maybe also in this case two unit fractions are enough 5?
For n ≠ 6l + 1 such fractions can always be written as a sum of only two
different unit fractions.
2
3
and , respectively
n
n
have a representation as a sum of two different unit fractions could also be
tackled by students because the strategy “extract the biggest unit fraction” is
sufficient in the general cases, and this strategy is relatively clear.
General reasoning and proofs play a big role when dealing with this
topic. In all cases the proofs are short and relatively easy to understand.
The question of which fractions of the form
5 The following holds (proof see below): There is no representation of 3 as a sum of two
n
(different) unit fractions if and only if all prime factors of n are of the form 6l + 1 .
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4. FRACTIONS AS A SUM OF TWO DIFFERENT
UNIT FRACTIONS
4.1 Unit fractions
Several fractions have different representations as a sum of two
1 1 2 1 1
different unit fractions, above we had the example, + = = + .
6 18 9 5 45
1
But also considering the unit fraction
we have the same phenomenon of
6
different representations: besides the representation found above (according
1 1 1
to the principle “biggest unit fraction extracted”) = +
one can find,
6 7 42
1 1 1 1 1
by trial, the representations
+ = = + . The maximum
9 18 6 10 15
denominator here is smaller than in the earlier representation.
We now address the question: If a fraction has at least one
representation as a sum of two different unit fractions, how can we find all
such representations?
We establish a criterion for which fractions have such a
representation.
In our first attempt we stick to unit fractions, we know already that
they have such a representation. But which of them have more than one?
How can one find all of them?
Substantial teacher help may be needed to answer this question.
1 1 1
From the relation = + with x ≠ y it follows immediately that x > n
n x y
and y > n , thus we can write x = n + a and y = n + b with a ≠ b . We
assume, without loss of generality, a < b . Now we have (see [5], p. 44):
1
1
1
=
+
⇔ n 2 + na + nb + ab = n 2 + nb + n 2 + na ⇔ n 2 = ab
n n+a n+b
Theorem 3: Every pair (a,b) = (divisor, complementary divisor) of n2
1
1
1
.
yields a solution of =
+
n n+ a n+ b
As above, assume a < b , because a = b would bring the same unit
fractions and for a > b only the order of the unit fractions would be another
one.
For all n one has the possible “solutions” a = 1 and b = n 2 : With
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these we get the representation from above
1
1
1
.
=
+
n n + 1 n
+ n2
n⋅( n +1)
1
, we get that the pairs (a, b ) , with
6
>a
a < b and a ⋅ b = 36 , are: (1, 36); (2,18); (3,12); (4, 9) . Thus we have the
1 1 1 1 1 1 1 1
representations: + , + , + ,
+ . This means that in the
7 42 8 24 9 18 10 15
representations at the beginning of this section, there was only one missing:
1 1
+ .
8 24
With this procedure we found a method guaranteed to get all the
representations of a unit fraction as a sum of two different unit fractions.
In the example from above,
4.2 FRACTIONS OF THE FORM m
n
The general case can be treated in a similar way and should not be
too difficult for students. The following questions concerning the fractions
m
could be given to students for individual work if the method and the
n
1
results concerning are known:
n
m
• Which fractions
have a representation as a sum of two different unit
n
fractions?
m
• How can one find all such representations belonging to a fraction ?
n
Using a similar method as above, we can answer both questions. We
are looking for a criterion which allow us to quickly decide whether or not a
13
certain fraction, e.g.
, has such a representation. If yes, we are interested
28
in a method to find all such representations. We stick to positive fractions
m 3
and it is a priori clear that only fractions
≤ can be meant because the
n 2
1
1
biggest two unit fractions are and .
1
2
We take the analogous approach
m 1 1
(4)
= + ,
n x y
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where again (without loss of generality) x < y with m, n, x, y ∈ . We
1
1 1
rewrite (4) to
= + and can read off:
n/m x y
n
n+a
n
n+b
x>
⇒ x=
∈ and y >
⇒ y=
∈ , with 1 ≤ a < b
m
m
m
m
Therefore it is clear: m | (n + a ) and m | (n + b) .
Then (4) implies:
m
1
1
m
m
m
1
1
1
=
+
⇔
=
+
⇔
=
+
⇔ n 2 = ab
n n+a n+b
n n+a n+b
n n+a n+b
m
m
We can formulate the following criterion:
•
•
m 3
≤ has a representation as a sum of two different unit
n 2
fractions if and only if there exist pairs (a, b) = (divisor, complementary
A fraction 0 <
divisor of n 2 ) with the property m | (n + a ) and m | (n + b) .
For the number of such representations the following holds:
m
Every such pair (a, b ) yields a representation of
as a sum of two
n
>a
m 1 1
1
1
different unit fractions: = + =
+
n
+
a
n
+b
n x y
m
m
•
5. FIVE EXAMPLES
13
We now focus on the above example
: The prime factorisation of
28
282 = 784 is: 282 = 24 ⋅ 7 2 . This we use to establish a table for the
possible values of a and b:
a
1
2
4
7
8 14 16
b > a 784 392 196 112 98 56 49
If we examine the divisibilities 13 | (28 + a ) and 13 | (28 + b) , we can
13
see that no value of a in this table yields 13 | (28 + a ) . Therefore
28
has no representation as a sum of two different unit fractions.
In the next two examples the denominator is of the form 6l+1 and
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the numerator is 3 – this was the case in which we could not make the
decision whether we will need three or only two unit fractions. In the first
case we will see that there is no representation as a sum of two different unit
fractions (one would need three) in the second case there will be such a
representation, even a unique one.
•
•
3
. Here n 2 = 49 with the only possible pair 6
7
(1, 49) = (divisor, complementary divisor of 49 ). But 3 is not a
3
divisor of 7 + 1 = 8 (also not of 7 + 49 = 56). Thus
has no such
7
representation as a sum of two different unit fractions.
We take the fraction
Now try the fraction
3
. The possible divisors of 252 = 625 = 54
25
are:
a
b>a
1
625
5
125
The first pair (a, b) = (1, 625) brings no solution because 3 is not a
divisor of 25 + 1 = 26 (also not of 25 + 625 = 650). But from the second
pair (a, b) = (5,125) we get a solution:
3
1 1
3 | ( 25 + 5 ) and 3 | ( 25 + 125 ) and therefore:
= + ; we also
25 10 50
30
150
know from our investigations that other solutions cannot exist.
•
Using the fraction
divisors
of
3
one finds several solutions. Considering the
14
one
can
find:
142 = 196 = 22 ⋅ 7 2
a
1
2
4
7
b>a
196
98
49
28
The second pair (a, b) = (2,98) does not yield a solution because 3 is not
a divisor of 14 + 2 = 16. But the other three pairs yield solutions:
6
We assume a < b .
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 14 + 1 = 15
14 + 196 = 210


We can find here 3 | 14 + 4 = 18 and 3 |  14 + 49 = 63 ,
14 + 7 = 21
 14 + 28 = 42


thus we get here three different representations as a sum of two different
3 1 1
3 1 1
3 1 1
unit fractions:
= + ,
= + ,
= +
14 5 70 14 6 21 14 7 14
•
3
2
and respectively one recognises
10
9
1 1
3 1 1
and
that we already had all solutions earlier +
= = +
4 20 10 5 10
1 1 2 1 1 2
( would also have the “trivial” solution with
+ = = +
6 18 9 5 45 9
2 1 1
equal unit fractions: = + ).
9 9 9
If we apply our method to
Students could also try to write a computer program which has the
following contents: Running the program with a fraction, it should either say
that there exists no such a representation as a sum of two different unit
fractions or list all such representations.
6. A FURTHER THEOREM AND UP TO NOW
UNSOLVED PROBLEMS
One of the central messages of the paper is that when dealing with
this topic one is very close to mathematical research: unsolved problems;
which can be seen in this section. With the help of the above criterion one
can give reasons for the statement in footnote 5, above:
Theorem 4: 3 has a representation as a sum of two different unit
n
fractions ⇔ Not all prime factors of n are of the form 6l + 1
Proof: ( ⇒ ): We first state an important insight: If two numbers of the form
6l + 1 are multiplied the result (product) is also of this form:
(6l1 + 1)(6l2 + 1) = 6(6l1l2 + l1 + l2 ) + 1
3
If
has a representation as a sum of two different unit fractions
n
then – following the criterion above – there must exist a factorisation
n 2 = a ⋅ b with 3 | (n + a ) and 3 | (n + b) . If one assumes (indirectly) that all
prime factors of n are of the form 6l + 1 then also n itself would have to be
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of this form 6i + 1 and also all possible divisors a, b of n 2 would have to be
of the form 6k + 1 . Thus n + a = 6i + 1 + 6k + 1 = 6(i + k ) + 2 and this would
surely not be divisible by 3. We have arrived at a contradiction to 3 | (n + a )
assuming that all prime factors of n are of the form 6l + 1 , that means we
have shown that the assumption is wrong.
Here the principle of “indirect proofs” (proof by contradiction), can
be used and repeated in an elementary way.
( ⇐ ): If n has not only prime factors of the form 6l + 1 then n must have
prime factor 2 or 3 or one of the form 6l − 1 . In all three cases one can
guarantee that there exists such a representation:
3 1
• n = 3n1 : In this case we have =
and we already know that unit
n n1
fractions have a representation as a sum of two different unit
fractions.
• n = 2n1 : We are looking for factors a ⋅ b = n 2 = 4n12 with
•
3 | (2n1 + a ) and 3 | (2n1 + b) . With a = n1 and b = 4n1 these
requirements of divisibility are fulfilled.
n = (6l − 1) ⋅ n1 : We are looking for factors a ⋅ b = n 2 = (6l − 1) 2 n12
with 3 | [(6l − 1) ⋅ n1 + a] and 3 | [(6l − 1) ⋅ n1 + b] . With a = n1 and
n
n
2
b = (6 − 1) n1 these requirements of divisibility are fulfilled.
Of course, this proof cannot be expected of most students, but this
proof should not be missed when dealing with this topic: The phenomenon
of being near to unsolved problems can be seen best in number theory
because the corresponding questions or conjectures can be formulated in a
way that promotes understanding (e.g., Fermat’s last theorem, Goldbach
conjecture, etc.). In some fields of mathematics the open questions usually
cannot be understood by non mathematicians, they mostly do not know what
the formulations of the problems are about. In number theory this is not the
case! Quite the contrary, students have participated in solving the problem
with numerator 3, they are – so to speak – deeply involved, but for
numerator 4 and 5 the corresponding questions are still unsolved problems!
There are two relatively famous conjectures in the realm of Erdös,
Straus, and Sierpinski, which remain – so far as we know – unsolved:
4 1 1 1
• For all n ∈ , n > 1 there exist a, b, c ∈ with
= + +
n a b c
This conjecture was validated for all natural numbers 1 < n ≤ 1014 that
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means one has found a solution. But (as it is clear in mathematics): This
is by far not a proof that the conjecture holds for all n ∈ , n > 1 .
5 1 1 1
• For all n ∈ , n > 1 there exist a, b, c ∈ with = + +
n a b c
3
We have proved that
for n > 1 has always a representation of at
n
most three different unit fractions. If and only if n contains only prime
factors of the form 6l + 1 one needs the maximum number of 3 different unit
fractions, in all other cases we can manage it with not more than two.
REFERENCES
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3. Yiu-Kwong Man, "Modified Golomb algorithm for computing unit fraction expansions",
International Journal of Mathematics Education and Science. Technology, Vol.
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