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Hollow Cubes investigation
Hollow cubes like the picture above [reference] are an extension of the hollow squares investigation done previously. This time we can imagine a 3 dimensional stack of soldiers, and so try to work out which numbers of soldiers can be arranged into hollow cubes.
Therefore what we need to find is what numbers can be formed from a3-b3
Python code
We can write some Python3 code to find this out (this can be run here):
for k in range(1,200):
for a in range(0, 100):
for b in range(0,100):
if a**3-b**3 == k :
print(k,a,b)
This gives the following: (the first number is the number of soldiers and the 2 subsequent numbers are the 2 cubes).
1 1 0
7 2 1
8 2 0
19 3 2
26 3 1
27 3 0
37 4 3
56 4 2
61 5 4
63 4 1
64 4 0
91 6 5
98 5 3
117 5 2
124 5 1
125 5 0
127 7 6
152 6 4
169 8 7
189 6 3
We could perhaps investigate any patterns in these numbers, or explore how we can predict when a hollow cube has more than one solution. I’ll investigate which numbers can be written as both a hollow square and also a hollow cube.
Hollow squares and hollow cubes
list1=[]
for a in range(2, 50):
for b in range(2,50):
if a**2-b**2 !=0:
if a**2-b**2 > 0:
list1.append(a**2-b**2)
list2=[]
for j in list1:
for c in range(2,50):
for d in range(2,50):
if c**3-d**3 == j:
list2.append(c**3-d**3)
print(list2)
This returns the following numbers which can all be written as both hollow squares and hollow cubes.
[56, 91, 19, 117, 189, 56, 208, 189, 217, 37, 279, 152, 117, 448, 513, 504, 448, 504, 387, 665, 504, 208, 875, 819, 936, 817, 61, 999, 988, 448, 728, 513, 189, 1216, 936, 784, 335, 469, 1323, 819, 1512, 1352, 1197, 992, 296, 152, 1519, 1512, 1197, 657, 1664, 1323, 1647, 1736, 1701, 1664, 936, 504, 2107, 1387, 1216, 1027, 91, 2015, 279, 2232]
Hollow squares, cubes and hypercubes
Taking this further, can we find any number which can be written as a hollow square, hollow cube and hollow hypercube (4 dimensional cube)? This would require our soldiers to be able to be stretch out into a 4th dimensional space – but let’s see if it’s theoretically possible.
Here’s the extra code to type:
list1=[]
for a in range(2, 200):
for b in range(2,200):
if a**2-b**2 !=0:
if a**2-b**2 > 0:
list1.append(a**2-b**2)
list2=[]
for j in list1:
for c in range(2,200):
for d in range(2,200):
if c**3-d**3 == j:
list2.append(c**3-d**3)
print(list2)
for k in list2:
for e in range(2,200):
for f in range(2,200):
if k == e**4-f**4:
print(k)
Very pleasingly this does indeed find some solutions:
9919: Which can be formed as either 1002-92 or 223-93 or 104-34.
14625: Which can be formed as either 1212-42 or 253-103 or 114-24.
Given that these took some time to find, I think it’ll require a lot of computer power (or a better designed code) to find any number which is a hollow square, hollow cube, hollow hypercube and hollow 5-dimensional cube, but I would expect that there is a number out there that satisfies all criteria. Maybe you can find it?
Waging war with maths: Hollow squares
The picture above [US National Archives, Wikipedia] shows an example of the hollow square infantry formation which was used in wars over several hundred years. The idea was to have an outer square of men, with an inner empty square. This then allowed the men in the formation to be tightly packed, facing the enemy in all 4 directions, whilst the hollow centre allowed the men flexibility to rotate (and also was a place to hold supplies). It was one of the infantry formations of choice against charging cavalry.
So, the question is, what groupings of men can be arranged into a hollow square? This is a current Nrich investigation, so I thought I’d do a mini-investigation on this.
We can rethink this question as asking which numbers can be written as the difference between 2 squares. For example in the following diagram (from the Nrich task Hollow Squares)
We can see that the hollow square formation contains a larger square of 20 by 20 and a smaller hollow square of 8 by 8. Therefore the number of men in this formation is:
202-82 = 336.
The first question we might ask therefore is how many numbers from 1-100 can be written as the difference between 2 squares? These will all be potential formations for our army.
I wrote a quick code on Python to find all these combinations. I included 0 as a square number (though this no longer creates a hollow square, rather just a square!). You can copy this and run it in a Python editor like Repl.it.
for k in range(1,50):
for a in range(0, 100):
for b in range(0,100):
if a**2-b**2 == k :
print(k,a,b)
This returned the following results:
1 1 0
3 2 1
4 2 0
5 3 2
7 4 3
8 3 1
9 3 0
9 5 4
11 6 5
12 4 2
13 7 6
15 4 1
15 8 7
16 4 0
16 5 3
17 9 8
19 10 9
20 6 4
21 5 2
21 11 10
23 12 11
24 5 1
24 7 5
25 5 0
25 13 12
27 6 3
27 14 13
28 8 6
29 15 14
31 16 15
32 6 2
32 9 7
33 7 4
33 17 16
35 6 1
35 18 17
36 6 0
36 10 8
37 19 18
39 8 5
39 20 19
40 7 3
40 11 9
41 21 20
43 22 21
44 12 10
45 7 2
45 9 6
45 23 22
47 24 23
48 7 1
48 8 4
48 13 11
49 7 0
49 25 24
Therefore we can see that the numbers with no solutions found are:
2,6,10,14,18,22,26,30,34,38,42,46,50
which are all clearly in the sequence 4n-2.
Thinking about this, we can see that this can be written as 2(2n-1) which is the product of an even number and an odd number. This means that all numbers in this sequence will require an odd factor in each of their factor pairs:
eg. 50 can be written as 10 (even) x 5 (odd) or 2 (even) x 25 (odd) etc.
But with a2-b2 = (a+b)(a-b), due to symmetries we will always end up with (a+b) and (a-b) being both even or both odd, so we can’t create a number with a factor pair of one odd and one even number. Therefore numbers in the sequence 4n-2 can’t be formed as the difference of 2 squares. There are some nicer (more formal) proofs of this here.
A battalion with 960 soldiers
Next we are asked to find how many different ways of arranging 960 soldiers in a hollow square. So let’s modify the code first:
for a in range(0, 1000):
for b in range(0,1000):
if a**2-b**2 == 960 :
print(a,b)
Which gives us the following solutions:
31 1
32 8
34 14
38 22
46 34
53 43
64 56
83 77
122 118
241 239
General patterns
We can notice that when the number of soldiers is 1,3,5,7,9,11 (2n-1) we can always find a solution with the pair n and n-1. For example, 21 can be written as 2n-1 with n = 11. Therefore we have 10 and 11 as our pair of squares. This works because 112-102 = (11+10)(11-10) returns the factor pair 21 and 1. In general it always returns the factor pair, 2n-1 and 1.
We can also notice that when the number of soldiers is 4,8,12,16,20 (4n) we can always find a solution with the pair n+1 and n-1. For example, 20 can be written as 4n with n = 5. Therefore we have 6 and 4 as our pair of squares. This works because 62-42 = (6+4)(6-4) returns the factor pair 10 and 2. In general it always returns the factor pair, 2n and 2.
And we have already shown that numbers 2,6,10,14,18,22 (4n-2) will have no solution. These 3 sequences account for all the natural numbers (as 2n-1 incorporates the 2 sequences 4n-3 and 4n-1).
So, we have found a method of always finding a hollow square formation (if one exists) as well as being able to use some computer code to find other possible solutions. There are lots of other avenues to explore here – could you find a method for finding all possible combinations for a given number of men? What happens when the hollow squares become rectangles?
IB Maths Resources from Intermathematics 