Sum of Cubes Equals Square of Sum(SCESS)
Matthew Fan
Zhang Jing Wen
December 18, 2014
1
Main Problem
The main problems in this project we focused on are related to this equation:
a31 + a32 + ... + a3n = (a1 + a2 + ... + an )2
where all the ai ’s are positive integers.
2
Basic Identity
Let us observe a pattern:
13 = 12
13 + 23 = (1 + 2)2
13 + 23 + 33 = (1 + 2 + 3)2
13 + 23 + 33 + 43 = (1 + 2 + 3 + 4)2
The statement 13 + 23 + ... + n3 = (1 + ... + n)2 is true and can be proven
by mathematical induction (proof not included here).
3
A generalisation of the previous identity
Here are a group of solutions:
{2, 2}
{2, 2, 4, 4}
{2, 2, 4, 4, 6, 6}
also,
{3, 3, 3}
{3, 3, 3, 6, 6, 6}
1
Proposition 3.1. {n, ..., n, 2n, ..., 2n, ..., n2 , ..., n2 }, is also a multi-set that sat| {z } | {z }
| {z }
n
n
n
isfies the original equation.
Proof.
!2
n
X
ai
= (n × n + n × 2n + · + n × n2 )2 = n4 (1 + 2 + · · · + n)2
i=1
n
X
a3i = n × n3 + n × (2n)3 + · · · + n × (n2 )3 = n4 (13 + 23 + · · · + n3 ).
i=1
Since (1 + 2 + · · · + n)2 = 13 + 23 + · · · + n3 , {n, ..., n, 2n, ..., 2n, ..., n2 , ..., n2 }
| {z } | {z }
| {z }
n
n
n
satisfies the original equation.
4
A Few Definitions
Let us called a multi-set that satisfies the original equation a good set. So
{1, 2, 3, 4, 5} is a good set. Note that we changed the term multi-set to set for
convenience.
Let A and B be good sets such that set A = {a1 , . . . , am } and set B =
{b1 , . . . , bn } The product set P of sets A and B is the set{a1 b1 , . . . , am bn }.For
example, the product set of {1, 2, 3} and {1, 2} is {1, 2, 3, 2, 4, 6}
Proposition 4.1. A product set is a good set.
Proof. a31 b31 + · · · + a3m b3n = (a31 + · · · + a3m )(b31 + · · · + b3n ) = (a1 + · · · + am )2 (b1 +
· · · + bn )2 = (a1 b1 + · · · + am bn )2
5
Distinct elements
It has been proven by Edward Barbeau and Samer Seraj that if all the elements
in a good set is distinct, then it must be the set with elements {1, 2, ..., n}.
6
Mason’s sets of solutions
Proposition 6.1. Let positive integer k have the set of factors {a1 , ..., an }. Let
bi denote the number of factors of ai , then {b1 , ..., bn } is a good set (elements
are non-necessarily distinct).
This proposition has been proved by John Mason.
For example, let n = 12. Then the set of factors are {1, 2, 3, 4, 6, 12}. And
the set of bi is {1, 2, 2, 3, 4, 6}, which is a product set of {1, 2} and {1, 2, 3} and
hence is a good set.
2
Note: the following proof has the same idea as the proof given in [2].
Proof.
We will directly evaluate
n
X
b3i and
i=1
n
X
!2
bi
to show that they are equal.
i=1
Let k = pq11 pq22 . . . pqnn .
Then {a1 , a2 , . . . , an } is the set which contains all factors of k, which means
that
{a1 , a2 , . . . , an } = {pr11 pr22 . . . prnn |0 ≤ ri ≤ qi }.
If ai = pr11 pr22 . . . prnn , then bi = (1 + r1 )(1 + r2 ) . . . (1 + rn ). Hence
{b1 , b2 . . . , bn } = {(1 + r1 )(1 + r2 ) . . . (1 + rn )|0 ≤ ri ≤ qi }.
If we fix the value of r2 , . . . , rn and vary the value of r1 from 0 to q1 , we see
that every different input of r1 gives one different output, which is one of the
elements in the set {b1 , . . . , bn }.
" q
#
1
X
The sum of these outputs is equal to
(1 + r1 ) (1 + r2 ) . . . (1 + rn ).
r1 =0
Since every ri is allowed to vary from 0 to qi , we have
!2 "r =q
#2
n
i
i
X
X
bi
=
(1 + r1 )(1 + r2 ) . . . (1 + rn )
ri =0
i=1
"
=
q1
X
#2 "
(1 + r1 )
r1 =0
q2
X
#2
(1 + r2 )
"
...
r2 =0
#2
qn
X
(1 + rn )
rn =0
= [1 + · · · + (q1 + 1)]2 [1 + · · · + (q2 + 1)]2 . . . [1 + · · · + (qn + 1)]2 .
Similarly, we have
! "r =q
#
n
i
i
X
X
3
3
3
3
bi =
(1 + r1 ) (1 + r2 ) . . . (1 + rn )
ri =0
i=1
"
=
q1
X
#"
(1 + r1 )
r1 =0
3
q2
X
#
(1 + r2 )
r2 =0
3
"
...
qn
X
#
3
(1 + rn )
rn =0
= [1 + · · · + (q1 + 1)3 ][1 + · · · + (q2 + 1)3 ] . . . [1 + · · · + (qn + 1)3 ].
!2
n
n
X
X
3
2
3
Since [1 + · · · + (qi + 1)] = [1 + ... + (qi + 1) ], we have
bi and
bi
i=1
are equal.
3
i=1
7
Other sets of solutions
{a1 , a2 , . . . , an } = {3k, . . . , 3k , 3k + 3, 3k + 6, . . . , 6k − 3}
| {z }
3k
Proof.
[3k(3k) + 3k + 3 + ... + 6k − 3]2 = [9k 2 + 3((1 + ... + 2k − 1) − (1 + ... + k))]2
2
(2k − 1)(2k) k(k + 1)
2
2
= 3 3k + 3
−
2
2
2
k(k − 1)
= 32 3k 2 + 3
2
2
k(k − 1)
= 34 k 2 +
2
2
k−1
4 2
=3 k k+
2
2
3k − 1
= 34 k 2
2
3k(3k)3 + (3k + 3)3 + ... + (6k − 3)3 = 33 (3k 4 + [(13 + ... + (2k − 1)3 ) − (1 + ... + k 3 (])
(2k − 1)(2k)
k(k + 1)
3
4
= 3 3k +
−
2
2
2
2
16k − 16k + 4 − k − 2k − 1
3
4
2
= 3 3k + k
4
2
5k − 6k + 1
= 34 k 2 k 2 +
4
2
9k
−
6k
+
1
4 2
=3 k
4
2
3k − 1
= 34 k 2
2
So they are equal.
{a1 , a2 , . . . , an } = {3k, . . . , 3k , 3k + 3, 3k + 6, . . . , 6k}
| {z }
3k+1
4
Proof.
[(3k + 1)(3k) + 3k + 3 + ... + 6k]2 = (9k 2 + 3[(1 + ... + 2k) − (1 + ... + (k − 1))])2
2
k(k + 1)
2
2
= 3 3k + 3
2
k(k + 1) 2
)
= 34 (k 2 +
2
k+1 2
= 34 k 2 (k +
)
2
3k + 1 2
= 34 k 2 (
)
2
(3k + 1)(3k)3 + (3k + 3)3 + ... + (6k)3 = 33 (3k 4 + [(13 + ... + (2k)3 ) − (1 + ... + (k − 1)3 )])
16k 2 + 16k + 4 − k 2 + 2k − 1
))
4
5k 2 + 6k + 1
= 34 k 2 (k 2 +
)
4
9k 2 + 6k + 1
= 34 k 2 (
)
4
3k + 1 2
= 34 k 2 (
)
2
= 33 (3k 4 + k 2 (
So they are equal.
{a1 , a2 , . . . , an } = {1, 2, 2, 3, 4, . . . , n − 2, n}
Proof.
(1 + 2 + 2 + 3 + · · · + (k − 2) + k)2 = (1 + 2 + 3 + ... + k − (k − 3))2
= (1 + 2 + 3 + ... + k)2 − (k − 3)(2(1 + ... + k) − (k − 3))
= 13 + ... + k 3 + (k − 3)2 − (k − 3)(k(k + 1))
= 13 + ... + k 3 + k 2 − 6k + 9 − k 3 + 2k 2 + 3k
= 13 + ... + k 3 − (k 3 − 3k 2 + 3k − 1) + 8
= 13 + ... + k 3 − (k − 1)3 + 23
{a1 , a2 , . . . , an } = {2, 2n, . . . , 2n, 2n + 2, . . . , 2n + 2}.
| {z } |
{z
}
n
n+1
5
Proof.
(2 + n(2n) + (n + 1)2(n + 1))2 = 22 (1 + n2 + n2 + 2n + 1)2
= 42 (n2 + n + 1)2
= 42 (n4 + 2n3 + 3n2 + 2n + 1)
= 8n4 + (8n4 + 32n3 + 48n2 + 32n + 8) + 8
= n(2n)3 + (n + 1)[2(n + 1)]3 + 23
8
Upper bound on the sum
We have some good sets from n = 1 to n = 5
1:
2:
3:
4:
5:
{1}
{1, 2}, {2, 2}
{1, 2, 3}, {3, 3, 3}
{1, 2, 2, 4}, {1, 2, 3, 4}, {2, 2, 4, 4}, {4, 4, 4, 4}
{1, 2, 2, 3, 5}, {1, 2, 3, 4, 5}, {3, 3, 3, 3, 6}, {3, 3, 3, 4, 6}, {5, 5, 5, 5, 5}
From here, we see no particular pattern on the largest possible element in a
good set for n. An observation that we can make from these good sets is that
the sum of the elements of a good set with n elements never exceeds n2 (when
all the elements are n). This will be proved later.
9
Proving the upper bound on the sum
Suppose that a1 + · · · + an > n2 , then let it be of the form n2 + k, where k > 0.
Then (a1 + · · · + an )2 = (n2 + k)2 = n4 + 2n2 k + k 2 .
Proof. By Cauchy-Schwarz inequality, we have
1) (a31 + · · · + a3n )(a1 + · · · + an ) ≥ (a21 + · · · + a2n )2
2) (a21 + · · · + a2n ) (1 + · · · + 1) ≥ (a1 + · · · + an )2 .
|
{z
}
n
Hence,
2
(a31 + · · · + a3n )(a1 + · · · + an )(1 + · · · + 1) ≥ (a21 + · · · + a2n )2 (1 + · · · + 1)
|
{z
}
|
{z
}
n
n
4
≥ (a1 + · · · + an ) .
6
2
This means that
(n2 + k)3
− (n2 + k)2
n2
k3
= n4 + 3n2 k + 3k 2 + 2 − (n4 + 2n2 k + k 2 )
n
k3
2
2
= n k + 2k + 2
n
>0
a31 + · · · + a3n − (a1 + · · · + an )2 ≥
Hence, if a1 + · · · + an > n2 , then a31 + · · · + a3n > (a1 + · · · + an )2 and there can
be no solutions.
The bound of n2 is achieved when a1 = . . . = an = n.
10
Lower bound on the sum
Here are the minimum values of the sum of elements in a good set as n varies.
Number
1
2
3
4
5
6
7
Minimum value of sum of elements
1
3
6
9
13
12
22
There seems to be no specific patterns for the lower bound on the sum other
than the fact that it is generally increasing.
11
The most number of times the same element
appears in a good set.
{a1 , . . . , an } = {n, . . . , n} is a good set which have the same element appearing
n times. Besides the above set, what is the largest number of times that the
7
same element can appear in a good set with n elements?
n Largest number of appearances of an element in a n-element good set
1
None
2
1
3
1
4
2
5
4
6
5
7
6
8
6
9
8
10
9
11
10
12
11
From here we see that the appearances seem to range from n − 2 to n − 1.
A question that arises from this is the question of whether we can find a good
set with at most two distinct elements.
12
Upper bound on largest element
4
Proposition 12.1. The largest element of a good set does not exceed n 3 .
Number
2
3
4
5
6
7
8
Largest element
2
3
4
6
7
9
10
4
n3
2.51984
4.32675
6.34960
8.54988
10.9027
13.3905
16
Note that we used an approximate value of x here.
As we can see, this bound is not a very tight bound for the largest element.
13
Better upper bound on largest element
Theorem 13.1. Let the largest element be an . Then anò x where x is the
(n−1)+ (n−1)2 +6x
solution of this equation in terms of n: ((n − 1)[
] + x)2 =
3
√
(n−1)+ (n−1)2 +6x 3
(n − 1)[
] + x3 .
3
8
Number
2
3
4
5
6
7
8
9
10
11
12
13
x
bxc An example of a good set with this bound
2.15470 2
(2, 2)
3.41935 3
(3, 3, 3)
4.77291 4
(4, 4, 4, 4)
6.20248 6
(6, 3, 3, 3, 3)
7.69915 7
(7, 6, 6, 5, 5, 4)
9.25630 9
(9, 6, 6, 6, 6, 6, 6)
10.8688 10
(10, 9, 6, 6, 6, 6, 6, 6)
12.5324 12
(12, 9, 6, 6, 6, 6, 6, 6, 6)
14.2438 14
(14, 9, 8, 8, 8, 8, 8, 8, 8, 8)
16
16
(16, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8)
17.7986 17
(17, 13, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9)
19.6373 19 (19, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 10)
Note that we used an approximate value of x here.
As we can see, this bound is a much better bound.
Proof. If we fix the value of a1 + · · · + an−1 , then an achieves its maximum value
when a31 + · · · + a3n−1 is minimal, which occurs when a1 = . . . = an−1 .
Hence, when an achieves maximum, [(n − 1)a1 + an ]2 = (n − 1)a31 + a3n .
So a3n achieves its maximum when f (a1 ) = [(n − 1)a1 + an ]2 − (n − 1)a31 achieves
its maximum, which is when f 0 (a1 ) = 0.
2
Then we have (n−1)(2a
√n +a1 (2n−3a1 −2)) = 0 =⇒ 3a1 +(2−2n)a1 −2an = 0
which gives a1 =
n−1+
(n−1)2 +6an
.
3
This results in the maximum√value of an being the positive, real
of
√ solution
2 +6x
(n−1)+ (n−1)2 +6x
(n−1)+
(n−1)
] + x)2 = (n − 1)[
]3 +
this equation: ((n − 1)[
3
3
x3 .
9
Number
14
15
16
17
18
19
20
21
22
23
x
21.5144
23.4281
25.3769
27.3594
29.3744
31.4209
33.4977
35.6039
37.7387
39.9012
bxc
21
23
25
27
29
31
33
35
17
39
As we can see, for n = 14, . . . , 23, we have bxc = 2n − 7.
We will show that {a1 , a2 . . . , an } = {n − 5, . . . , n − 5, n − 2, n − 2, 2n − 7}
{z
}
|
n−3
is a good set.
Proof. Let m = n − 2.
Then
(a1 + · · · + an )2 = [(n − 3)(n − 5) + 2(n − 2) + 2n − 7]2
= [(m − 1)(m − 3) + 2m + (2m − 3)]2
= (m2 )2
= m4 .
And
a31 + · · · + a3n = (m − 1)(m − 3)3 + 2m3 + (2m − 3)3
= (m − 1)(m3 − 9m2 + 27m − 27) + 2m3 + (8m3 − 36m2 + 54m − 27)
= m4 .
So they are equal and these are good sets.
10
14
Lower bound on the largest element
Here are the minimum values for the largest element when n varies.
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
The minimum value of the largest element of a good set
1
2
3
4
5
5
6
5
7
6
6
7
8
8
9
9
9
9
9
9
10
10
10
11
10
As we can see, the minimum value of the largest element in a good set is generally
increasing as n increase but there are a few outliers at 7 to 8, 9 to 10 and 24 to
25. Hence it is unlikely there is a good bound.
15
Future directions
As we go on, we may consider
1) whether we can create other groups of solution like John Mason
2) whether the number of good sets grow as n grows
3) whether we always have a good set that have the maximum value given from
the formula of the tight bound
4) what is the largest number of times a certain element can appear in the
set(besides the case where everything is n)
11
16
References
[1] John Mason, Generalising ‘Sum of cubes equal to squares of sums’, 2001.
[2] Edward Barbeau, Samer Seraj, University of Toronto, Sum of cubes is square
of sum,http://arxiv.org/pdf/1306.5257.pdf 2013
12