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The IB Maths Exploration Guide and the IB Maths Modelling and Statistics Exploration Guide are suitable for all IB students.  

They are both written by an IB teacher with an MSc. in Mathematics, 10 years experience teaching IB Standard and Higher Level and who has worked as an IB examiner.

PLEASE NOTE: This is not an automatic download I will send all orders received within 24 hours.

Resource Number 1

The Exploration Guide talks you through:

  1. An introduction to the essentials about the investigation,
  2. The new marking criteria,
  3. How to choose a topic,
  4. Examples of around 70 topics that could be investigated,
  5. Useful websites for use in the exploration,
  6. A student checklist for completing a good investigation,
  7. Common mistakes that students make and how to avoid them,
  8. General stats projects advice,
  9. A selection of some interesting exploration topics explored in more depth,
  10. Teacher advice for marking,
  11. Templates for draft submissions,
  12. Advice on how to use Geogebra, Desmos and Tracker in your exploration,
  13. Some examples of beautiful maths using Geogebra and Desmos.

Exploration Guide

A comprehensive 63 page pdf guide to help you get excellent marks on your maths investigation. [Will be emailed within the same day as ordered].

7.50 $

Resource Number 2

The Modelling and Statistics Guide talks you through various techniques useful for statistical and modelling explorations. Topics included are:

  1. Linear regression
  2. Quadratic regression
  3. Cubic regression
  4. Exponential regression
  5. Linearisation using log scales
  6. Trigonometric regression
  7. Pearson’s Product investigations: Height and arm span
  8. Binomial investigations: ESP powers
  9. Poisson investigations: Customers in a shop
  10. 2 sample t tests: Reaction times
  11. Paired t tests: Reaction times
  12. Chi Squared: Efficiency of vaccines
  13. Bernoulli trials: Polling confidence intervals
  14. Spearman’s rank: Taste preference of cola
  15. Sampling techniques and experiment design.

Modelling and Statistics Guide

A 60 page pdf guide full of advice to help with modelling and statistics explorations. [Will be emailed within the same day as ordered].

7.50 $

Resource Number 3

The Exploration Guide and the Modelling and Statistics Guide can be purchased together for a discount.

Exploration Guide AND the Modelling and Statistics Guide

Both guides included together for a discount. [Will be emailed within the same day as ordered].

10.50 $


The Martingale system

The Martingale system was first used in France in 1700s gambling halls and remains used today in some trading strategies.  I’ll look at some of the mathematical ideas behind this and why it has remained popular over several centuries despite having a long term expected return of zero.

The scenario

You go to a fair ground and play a simple heads-or-tails game.  The probability of heads is 1/2 and tails is also 1/2.  You place a stake of counters on heads.  If you guess correctly you win that number of counters.  If you lose, you double your stake of counters and then the coin is tossed again.  Every time you lose you double up your stake of counters and stop when you finally win.

Infinitely deep pockets model:

You can see that in the example above we always have a 0.5 chance of getting heads on the first go, which gives a profit of 1 counter.  But we also have a 0.5 chance of a profit of 1 counter as long as we keep doubling up our stake, and as long as we do indeed eventually throw heads.  In the example here you can see that the string of losing throws don’t matter [when we win is arbitrary, we could win on the 2nd, 3rd, 4th etc throw].  By doubling up, when you do finally win you wipe out your cumulative losses and end up with a 1 counter profit.

This leads to something of a paradoxical situation, despite only having a 1/2 chance of guessing heads we end up with an expected value of 1 counter profit for every 1 counter that we initially stake in this system.

So what’s happening?  This will always work but it requires that you have access to infinitely deep pockets (to keep your infinite number of counters) and also the assumption that if you keep throwing long enough you will indeed finally get a head (i.e you don’t throw an infinite number of tails!)

Finite pockets model:

Real life intrudes on the infinite pockets model – because in reality there will be a limit to how many counters you have which means you will need to bail out after a given number of tosses.  Even if the probability of this string of tails is very small, the losses if it does occur will be catastrophic –  and so the expected value for this system is still 0.

Finite pockets model capped at 4 tosses:

In the example above we only have a 1/16 chance of losing – but when we do we lose 15 counters.  This gives an expected value of:

Finite pockets model capped at n tosses:

If we start with a 1 counter stake then we can represent the pattern we can see above for E(X) as follows:

Here we use the fact that the losses from n throws are the sum of the first (n-1) powers of 2. We can then notice that both of these are geometric series, and use the relevant formula to give:

Therefore the expected value for the finite pockets model is indeed always still 0.

So why does this system remain popular?

So, given that the real world version of this has an expected value of 0, why has it retained popularity over the past few centuries?  Well, the system will on average return constant linear growth – up until a catastrophic loss.  Let’s say you have 100,000 counters and stake 1 counter initially.  You can afford a total of 16 consecutive losses.  The probability of this is only:

but when you do lose, you’ll lose a total of:

So, the system creates a model that mimics linear growth, but really the small risk of catastrophic loss means that the system still has E(X) = 0.  In the short term you would expect to see the following very simple linear relationship for profit:

With 100,000 counters and a base trading stake of 1 counter, if you made 1000 initial 1 counter trades a day you would expect a return of 1000 counters a day (i.e 1% return on your total counters per day).  However the longer you continue this strategy the more likely you are to see a run of 16 tails – and see all your counters wiped out.

Computer model

I wrote a short Python code to give an idea as to what is happening. Here I started 9 people off with 1000 counters each.  They have a loss limit of 10 consecutive losses.  They made starting stakes of 1 counter each time, and then I recorded how long before they made a loss of 10 tosses in a row.

For anyone interested in the code here it is:


The program returned the following results.  The first number is the number of starting trades until they tossed 10 tails in a row.  The second number was their new account value (given that they had started with 1000 counters, every previous trade had increased their account by 1 counter and that they had then just lost 1023 counters).

1338, 1315
1159, 1136
243, 220
1676, 1653
432, 409
1023, 1000
976, 953
990, 967
60, 37

This was then plotted on Desmos. The red line is the trajectory their accounts were following before their loss.  The horizontal dotted line is at y = 1000 which represents the initial account value.  As you can see 6 people are now on or below their initial starting account value.  You can also see that all these new account values are themselves on a line parallel to the red line but translated vertically down.

From this very simple simulation, we can see that on average a person was left with 884 counters following hitting 10 tails.  i.e below initial starting account.  Running this again with 99 players gave an average of 869.

999 players

I ran this again with 999 players – counting what their account value would be after their first loss.  All players started with 1000 counters.  The results were:

31 players bankrupt: 3%

385 players left with less than half their account value (less than 500): 39%

600 players with less than their original account value (less than 1000): 60%

51 players at least tripled their account (more than 3000): 5%

The top player ended up with 6903 counters after their first loss.

The average account this time was above starting value (1044.68).  You can see clearly that the median is below 1000 – but that a small number of very lucky players at the top end skewed the mean above 1000.

Second iteration

I then ran the simulation again – with players continuing with their current stake.  This would have been slightly off because my model allowed players who were bankrupt from the first round to carry on [in effect being loaned 1 counter to start again].  Nevertheless it now gave:

264 players bankrupt: 26%

453 players left with less than half their account value (less than 500): 45%

573 players with less than their original account value (less than 1000): 57%

95 players at least tripled their account (more than 3000): 10%

The top player ended up with 9583 counters after their second loss.

We can see a dramatic rise in bankruptcies – now over a quarter of all players.  This would suggest the long term trend is towards a majority of players being bankrupted, though the lucky few at the top end may be able to escape this fate.

Screen Shot 2020-04-08 at 1.09.39 PM

This carries on our exploration of projectile motion – this time we will explore what happens if gravity is not fixed, but is instead a function of time.  (This idea was suggested by and worked through by fellow IB teachers Daniel Hwang and Ferenc Beleznay).   In our universe we have a gravitational constant – i.e gravity is not dependent on time.  If gravity changed with respect to time then the gravitational force exerted by the Sun on Earth would lessen (or increase) over time with all other factors remaining the same.

Interestingly time-dependent gravity was first explored by Dirac and some physicists have tried to incorporate time dependent gravity into cosmological models.  As yet we have no proof that gravity is not constant, but let’s imagine a university where it is dependent on time.

Projectile motion when gravity is time dependent

Screen Shot 2020-04-10 at 2.12.03 PM

We can start off with the standard parametric equations for projectile motion. Here v is the initial velocity, theta is the angle of launch, t can be a time parameter and g is the gravitational constant (9.81 on Earth).  We can see that the value for the vertical acceleration is the negative of the gravitational constant.  So the question to explore is, what if the gravitational constant was time dependent?  Another way to think about this is that gravity varies with respect to time.

Linear relationship

If we have the simplest time dependent relationship we can say that:

Screen Shot 2020-04-08 at 1.01.41 PM

where a is a constant.  If a is greater than 0 then gravity linearly increases as time increases, if a is less than 0 than gravity linearly decreases as time increases.  For matters of slight convenience I’ll define gravity (or the vertical acceleration) as -3at.  The following can then be arrived at by integration:

Screen Shot 2020-04-10 at 2.17.28 PM

This will produce the following graph when we fix v = 10, a = 2 and vary theta:

Screen Shot 2020-04-08 at 1.09.39 PM

Now we can use the same method as in our Projectile Motion Investigation II to explore whether these maximum points lie in a curve.  (You might wish to read that post first for a step by step approach to the method).

Screen Shot 2020-04-08 at 1.44.34 PM

therefore we can substitute back into our original parametric equations for x and y to get:

Screen Shot 2020-04-08 at 1.46.02 PM

We can plot this with theta as a parameter.  If we fix v = 4 and a =2 we get the following graph:

Screen Shot 2020-04-08 at 1.33.06 PM

Compare this to the graph from Projectile Motion Investigation II, where we did this with gravity constant (and with v fixed as 10):

Screen Shot 2020-04-06 at 9.34.04 PM

The Projectile Motion Investigation II formed a perfect ellipse, but this time it’s more of a kind of egg shaped elliptical curve – with a flat base.  But it’s interesting to see that even with time dependent gravity we still have a similar relationship to before!

Inverse relationship

Let’s also look at what would happen if gravity was inversely related to time.  (This is what has been explored by some physicists).

In this case we get the following results when we launch projectiles (Notice here we had to use the integration by parts trick to integrate ln(t)).  As the velocity function doesn’t exist when t = 0, we can define v and theta in this case as the velocity and theta value when t = 1.

Screen Shot 2020-04-10 at 2.34.42 PM

Now we use the same trick as earlier to find when the gradient is 0:

Screen Shot 2020-04-10 at 2.37.22 PM

Substituting this back into the parametric equations gives:

Screen Shot 2020-04-10 at 2.42.57 PM

The ratio v/a will therefore have the greatest effect on the maximum points.

v/a ratio negative and close to zero:

v = 40, a = -2000, v/a = -0.02

Screen Shot 2020-04-10 at 2.52.57 PM

This gives us close to a circle, radius v, centred at (0,a).

v = 1, a = -10, v/a = -0.1

Screen Shot 2020-04-10 at 2.59.20 PM

Here we can also see that the boundary condition for the maximum horizontal distance thrown is given by x = v(e).

v/a ratio negative and large:

v = 40, a = -2, v/a = -20.

Screen Shot 2020-04-10 at 2.48.30 PM

We can see that we get an egg shape back – but this time with a flatter bulge at the top and the point at the bottom.  Also notice how quickly the scale of the shape has increased.

v/a ratio n/a (i.e a = 0)

Screen Shot 2020-04-10 at 3.07.17 PM

Here there is no gravitational force, and so projectiles travel in linear motion – with no maximum.

Envelope of projectiles for the inverse relationship

This is just included for completeness, don’t worry if you don’t follow the maths behind this bit!

Screen Shot 2020-04-15 at 10.09.59 AM

Screen Shot 2020-04-15 at 10.10.11 AM

Therefore when we plot the parametric equations for x and y in terms of theta we get the envelope of projectile motion when we are in a universe where gravity varies inversely to time.  The following graph is generated when we take v = 300 and a = -10.  The red line is the envelope of projectiles.

Screen Shot 2020-04-15 at 10.11.19 AM

A generalized power relationship

Lastly, let’s look at what happens when we have a general power relationship i.e gravity is related to (a)tn.  Again for matters of slight convenience I’ll look at the similar relationship -0.5(n+1)(n+2)atn.

Screen Shot 2020-04-10 at 2.21.52 PM

This gives (following the same method as above:

Screen Shot 2020-04-08 at 6.48.11 PM

Screen Shot 2020-04-08 at 6.48.46 PM

As we vary n we will find the plot of the maximum points.  Let’s take the velocity as 4 and a as 2.  Then we get the following:

When n = 0:

Screen Shot 2020-04-08 at 8.18.07 PM

When n = 1:

Screen Shot 2020-04-08 at 8.16.42 PM

When n =2:

Screen Shot 2020-04-08 at 8.18.21 PM

When n = 10:

Screen Shot 2020-04-08 at 8.18.44 PM

We can see the general elliptical shape remains at the top, but we have a flattening at the bottom of the curve.

When n approaches infinity:

Screen Shot 2020-04-08 at 8.26.45 PM

We get this beautiful result when we let n tend towards infinity – now we will have all the maximum points bounded on a circle (with the radius the same as the value chosen as the initial velocity.  In the graph above we have a radius of 4 as the initial velocity is 4. Notice too we have projectiles traveling in straight lines – and then seemingly “bouncing” off the boundary!

If we want to understand this, there is only going to be a very short window (t less than 1) when the particle can upwards – when t is between 0 and 1 the effect of gravity is effectively 0 and so the particle would travel in a straight line (i.e if the initial velocity is 5 m/s it will travel 5 meters. Then as soon as t = 1, the gravity becomes crushingly heavy and the particle falls effectively vertically down.

Classical Geometry Puzzle: Finding the Radius

This is another look at a puzzle from Mind Your Decisions.  The problem is to find the radius of the following circle:

We are told that line AD and BC are perpendicular and the lengths of some parts of chords, but not much more!  First I’ll look at my attempt to solve this.  It’s not quite as “nice” as the solution in the video as it requires the use of a calculator, but it still does the job.

Method 1, extra construction lines:

These are the extra construction lines required to solve this problem.  Here is the step by step thought process:

  1. Find the hypotenuse of triangle AGC.
  2. Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
  3. Therefore triangle AGC and GBD are similar, so length BG = 4.  We can now use Pythagoras to find length BD.
  4. We can find length CD by Pythagoras.
  5. Now we have 3 sides of a triangle, CDB.  This allows use to find angle BDC using the cosine rule.
  6. Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
  7. Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
  8. Now we have a right angled triangle BCE where we know both an angle and a side, so can use trigonometry to find the length of BE.
  9. Therefore the radius is approximately 4.03.

Method 2, creating a coordinate system

This is a really beautiful solution – which does not require a calculator (and which is discussed in the video above).  We start by creating a coordinate system based around point G at (0,0).  Because we have perpendicular lines we can therefore create coordinates for A, B and C.  We also mark the centre of the circle as (p,q).

First we start with the equation of a circle centre (p.q):

Next we create 3 equations by substituting in our coordinates:

Next we can do equation (3) – equation (1) to give:

Next we can substitute this value for p into equations (1) and (3) and equate to get:

Lastly we can substitute both values for p and q into equation (1) to find r:

We get the same answer as before – though this definitely feels like a “cleaner” solution.  There are other ways to solve this – but some of these require the use of equations you may not already know (such as the law of sines in a circumcircle, or the equation for perpendicular chords and radius).  Perhaps explore any other methods for solving this – what are the relative merits of each approach?

The Mordell Equation [Fermat’s proof]

Let’s have a look at a special case of the Mordell Equation, which looks at the difference between an integer cube and an integer square.  In this case we want to find all the integers x,y such that the difference between the cube and the square gives 2.  These sorts of problems are called Diophantine problems and have been studied by mathematicians for around 2000 years.  We want to find integer solution to:

First we can rearrange and factorise, using the property of imaginary numbers.

Next we define alpha and beta such that:

For completeness we can say that alpha and beta are part of an algebraic number field:

Next we use an extension of the Coprime Power Trick, which ensures that the following 2 equations have solutions (if our original equation also has a solution). Therefore we define:

We can then substitute our definition for alpha into the first equation directly above and expand:

Next we equate real and imaginary coefficients to give:

This last equation therefore requires that either one of the following equations must be true:

If we take the case when b = 1 we get:


If we take the case when b = -1 we get

Therefore our solution set is (a,b): (1,1), (1,-1), (-1,1), (-1,-1.  We substitute these possible answers into our definition for y to give the following:

We can then substitute these 2 values for y into the definition for x to get:

These therefore are the only solutions to our original equation.  We can check they both work:

We can see this result illustrated graphically by plotting the graph:

and then seeing that we have our integer solutions (3,5) and (3,-5) as coordinate on this curve.

This curve also clearly illustrates why we have a symmetrical set of solutions, as our graph is symmetrical about the x axis.

This particular proof was first derived by Fermat (of Fermat’s Last Theorem fame) in the 1600s and is an elegant example of a proof in number theory.  You can read more about the Mordell Equation in this paper (the proof above is based on that given in the paper, but there is a small mistake in factorization so that y = 7 and y = -7 is erroneously obtained)


Can you solve Oxford University’s Interview Question?

The excellent Youtube channel Mind Your Decisions is a gold mine for potential IB maths exploration topics.  I’m going to follow through my own approach to problem posed in the video. The problem is to be able to trace the movement of the midpoint of a ladder as it slides down a wall.  This has apparently been used as an Oxford interview question to test the ability to investigate novel problems.

Specific case

It’s normally a good idea to start with a specific case with some nice numbers, to see what happens.  So, I’ll choose a 3,4,5 triangle, where the ladder has a fixed length of 5 and has endpoints with coordinates (0,4) and (3,0).  The midpoint is given by ((0+3)/2, (4+0)/2) = (1.5, 2).

Next I imagine what would happen to the point (3,0) if the ladder slipped down the wall.  (3,0) would become (3+t,0) where t is a parameter.  Given that the length is fixed as 5, I can now find the new height of the ladder up the wall using Pythagoras:

The new height is given by:

Therefore the new midpoint is given by:

We can now define our curve parametrically:

Therefore we can make t the subject in the first equation to get an equation just in terms of x and y.

Therefore we can rearrange to get:

This is the equation of a circle centred at (0,0) with radius 2.5:

This graph therefore traces the movement of the midpoint of the ladder (note that when the ladder was vertical against the wall the midpoint would be 2.5 high hence the graph starts at (0,2.5).

The general case

Now we have worked through the maths for a specific case, the general case isn’t too much extra work.  For a triangle with base a and height b we would have the following midpoint coordinates:

This would lead to the following equation:

Which would rearrange to give the equation of the circle:

This is a circle centered at (0,0) with radius:

Another approach

This method is an alternative to the version above – this time using trigonometry.  We start with the triangle below:


and then let the ladder slide to get the following (as the angle will get smaller t will be negative):

We can then define the midpoint coordinates as:

We can then rearrange and square both sides to get the following:


We can then use the trig identity for cosine squared theta + sine squared theta = 1:

Which rearranges to give the same result as before:

So, there we go – we’ve passed an Oxford interview question with a couple of different methods!  The approach of first exploring the topic with a simple case is often a good starting point for these sorts of problems – as it allows you to gain an understanding of what is happening without getting too bogged down with variables.  You can watch the video for a quicker solution – are there any other ways of approaching this problem you can find?  How could this problem be modified?




Rational Approximations to Irrational Numbers

This year two mathematicians (James Maynard and Dimitris Koukoulopoulos) managed to prove a long-standing Number Theory problem called the Duffin Schaeffer Conjecture.  The problem is concerned with the ability to obtain rational approximations to irrational numbers.  For example, a rational approximation to pi is 22/7.  This gives 3.142857 and therefore approximates pi to 2 decimal places.  You can find ever more accurate rational approximations and the conjecture looks at how efficiently we can form these approximation, and to within what error bound.

Finding Rational Approximations for  pi

The general form of the inequality I want to solve is as follows:

Here alpha is an irrational number, p/q is the rational approximation, and f(q)/q can be thought of as the error bound that I need to keep my approximation within.

If I take f(q) = 1/q then I will get the following error bound:

So, the question is, can I find some values of q (where p and q are integers) such that the error bound is less than 1/(q squared)?

Let’s see if we can solve this for when our irrational number is pi, and when we choose q = 6.

We can see that this returns a rational approximation, 19/6 which only 0.02507… away from  pi. This is indeed a smaller error than 1/36.  We won’t be able to find such solutions to our inequality for every value of q that we choose, but we will be able to find an infinite number of solutions, each getting progressively better at approximating pi.

The General Case (Duffin Schaeffer Conjecture)

The general case of this problem states that there will be infinite solutions to the inequality for any given irrational number alpha if and only if the following condition holds:


We will have infinitely many solutions (with p and q as integers in their lowest terms) if and only if:

Here the new symbol represents the Euler totient.  You can read about this at the link if you’re interested, but for the purposes of the post we can transform into something else shortly!

Does f(q) = 1/q provide infinite solutions?

When f(q) = 1/q we have:

Therefore we need to investigate the following sum to infinity:

Now we can make use of an equivalence, which shows that:


Where the new symbol on the right is the Zeta function.  The Zeta function is defined as:

So, in our case we have s = 2.  This gives:

But we know the limit of both the top and the bottom sum to infinity.  The top limit is called the Harmonic series, and diverges to infinity. Therefore:

Whereas the bottom limit is a p-series with p=2, this is known to converge. In fact we have:

Therefore because we have a divergent series divided by a convergent one, we will have the following result:

This shows that our error bound 1/(q squared) will be satisfied by infinitely many values of q for any given irrational number.

Does f(q) = 1/(q squared) provide infinite solutions?

With f(q) = 1/(q squared) we follow the same method to get:

But this time we have:

Therefore we have a convergent series divided by a convergent series which means:

So we can conclude that f(q) = 1/(q squared) which generated an error bound of 1/(q cubed) was too ambitious as an error bound – i.e there will not be infinite solutions in p and q for a given irrational number.  There may be solutions out there but they will be rare.

Understanding mathematicians 

You can watch the Numberphile video where James Maynard talks through the background of his investigation and also get an idea what a mathematician feels like when they solve a problem like this!

When do 2 squares equal 2 cubes?

Following on from the hollow square investigation this time I will investigate what numbers can be written as both the sum of 2 squares, 2 cubes and 2 powers of 4. i.e a2+b2 = c3+d3 = e4+f4.

Geometrically we can think of this as trying to find an array of balls such that we can arrange them into 2 squares, or we can rearrange them and stack them to form 2 cubes, or indeed we can arrange them into 2 4-dimensional cubes. I’ll add the constraints that all of a,b,c,d,e,f should be greater than 1 and that the pair of squares or cubes (etc) must be distinct. Therefore we can’t for example have 2 squares the same size.

Infinite solutions

Let’s look at why we can easily find infinite solutions if the squares or cubes (etc) can be the same size.

We want to find solutions to:
a2+b2 = c3+d3 = e4+f4.

so we look at the powers 2,3,4 which have LCM of 12. Therefore if we choose powers with the same base we can find a solution. For example we chose to work with base 2. Therefore we choose

a = 26, b = 26, which gives 212+212
c = 24, d = 24, which gives 212+212
e = 23, f = 23, which gives 212+212

Clearly these will be the same. So we can choose any base we wish, and make the powers into the same multiples of 12 to find infinite solutions.

Writing some code

Here is some code that will find some other solutions:

for a in range(2, 200):
 for b in range(2,200):

for j in list1:
 for c in range(2,200):
  for d in range(2,200):
   if c**3+d**3 == j:

for k in list2:
 for e in range(2,200):
  for f in range(2,200):
   if k == e**4+f**4:

This returns the following solutions: 8192, 18737, 76832. Of these we reject the first as this is the solution 212+212 which we found earlier and which uses repeated values for the squares, cubes and powers of 4. The 3rd solution we also reject as this is formed by 14 4 + 14 4. Therefore the only solution up to 79202 (we checked every value up to and including 1992 + 1992) is:

18737 = 642+1212 = 173+243 = 114+84.

Therefore if we had 18,737 balls we could arrange them into 2 squares, a 64×64 square and a 121×121 square. Alternatively we could rearrange them into 2 cubes, one 17x17x17 and one 24x24x24. Or we could enter a higher dimensional space and create 2 tesseracts one with sides 11x11x11x11 and the other with 14x14x14x14.

With only 1 solution for around the first 80,000 numbers it looks like these numbers are quite rare – could you find another one? And could you find one that also satisfies g5+h5?

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 :

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

for a in range(2, 50):
 for b in range(2,50):
  if a**2-b**2 !=0:
   if a**2-b**2 > 0:

for j in list1:
 for c in range(2,50):
  for d in range(2,50):
   if c**3-d**3 == j:

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:

for a in range(2, 200):
 for b in range(2,200):
  if a**2-b**2 !=0:
   if a**2-b**2 > 0:

for j in list1:
 for c in range(2,200):
  for d in range(2,200):
   if c**3-d**3 == j:

for k in list2:
 for e in range(2,200):
  for f in range(2,200):
   if k == e**4-f**4:

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?

Ramanujan’s Taxi Cabs and the Sum of 2 Cubes

The Indian mathematician Ramanujan (picture cite: Wikipedia) is renowned as one of great self-taught mathematical prodigies.  His correspondence with the renowned mathematician G. H Hardy led him to being invited to study in England, though whilst there he fell sick.  Visiting him in hospital, Hardy remarked that the taxi that had brought him to the hospital had a very “rather dull number” – number 1729.  Ramanujan remarked in reply, ” No Hardy, it’s a very interesting number!  It’s the smallest number expressible as the sum of 2 cubes in 2 different ways!”

Ramanujan was profoundly interested in number theory – the study of integers and patterns inherent within them.  The general problem referenced above is finding integer solutions to the below equation for given values of A:

In the case that A = 1729, we have 2 possible ways of finding distinct integer solutions:

The smallest number which can be formed through 3 distinct (positive) integer solutions to the equation is A = 87, 539, 319.

Although this began as a number theory problem it has close links with both graphs and group theory – and it is from these fields that mathematicians have gained a deeper understanding as to the nature of its solutions.  The modern field of elliptical curve cryptography is closely related to the ideas below and provides a very secure method of encrypting data.

We start by sketching the graph of:

For some given integer value of A. We will notice that the graph has a line of symmetry around y = x and also an asymptote at y = -x.  If we plot:

We can see that both our integer solutions to this problem (1,12) and (9,10) lie on the curve:

Group theory

Groups can be considered as sets which follow a set number of rules with regards to operations like multiplication, addition etc.  Establishing that a set is a group then allows certain properties to be inferred.  If we can establish the following rules hold then we can create an Abelian group.  If we start with a set A and and operation Θ.

1) Identity. For an element e in A, we have a Θ e = a for all a in A.

(for example 0 is the identity element for the addition operation for the set of integers numbers. a+0 = a for all a in the real numbers).

2) Closure.  For all elements a,b in A, a Θ b = c, where c is also in A.

(For example with the addition operation, the addition of 2 integers numbers is still an integer)

3) Associativity. For all elements a,b,c in A, (a Θ b) Θ c = a Θ (b Θ c)

(For example with the addition operation, (1+2) + 3 = 1 + (2+3) )

4) Inverse.  For each a in A there exists a b in A such that a Θ b = b Θ a = e.  Where e is the identity.

(For example with the addition operation, 4+-4 = -4+4 = 0.  0 is the identity element for addition)

5) Commutativity. For all elements a,b in A, a Θ b = b Θ a

(For example with the addition operation 1+2 = 2+1).

As we have seen, the set of integers under the operation addition forms an abelian group.

Establishing a group

So, let’s see if we can establish a Abelian group based around the rational coordinates on our graph.  We can demonstrate with the graph:

We then take 2 coordinate points with rational coordinates (i.e coordinates that can be written as a fraction of integers).  In this case A (1,12) and B (9,10).

We then draw the line through A and B.  This will intersect the graph in a 3rd point, C (except in a special case to be looked at in a minute).

We then reflect this new point C in the line y = x, giving us C’.

In this case C’ is the point  (46/3, -37/3)

We therefore define addition (our operation Θ) in this group as:

A + B = C’.

(1,12) + (9,10) = (46/3, -37/3).

We now need to deal with the special case when a line joining 2 points on the curve does not intersect the curve again.  This will happen whenever the gradient of this line is -1, which will make it parallel to the graph’s asymptote y = -x.

In this case we affix a special point at infinity to the Cartesian (x,y) plane.  We define this point as the point through which all lines with gradient -1 intersect.  Therefore in our expanded geometry, the line through AB will intersect the curve at this point at infinity. Let’s call our special point  Φ.  Now we have a new geometry,  the (x,y) plane affixed with Φ.

We can now create an Abelian group.  For any 2 rational points P(x,y), Q(x,y) we will have:

1) Identity.  P + Φ = Φ + P = P

2) Closure.  P + Q = R. (Where R(x,y) is also a rational point on the curve)

3) Associativity. (P+Q) + R = P+(Q+R)

4) Inverse.  P + (-P) = Φ

5) Commutativity.  P+Q = Q+P

Understanding the identity

Let’s see if we can understand some of these.  For the identity, if we have a point A on the line and the point at infinity then this will contain the line with gradient -1.  Therefore the line between the point at infinity and A will intersect the curve again at B.  Our new point, B’ will be created by reflecting this point in the line y = x.  This gets us back to point A.  Therefore P + Φ = P as required.

Understanding the inverse

With the inverse of our point P(x,y) given as -P = (-x,-y) we can see that this is the reflection in the line y = x.  We can see that we we join up the 2 points reflected in the line y = x we will have a line with slope -1, which will intersect with the curve at our point at infinity.  Therefore P + (-P) = Φ.

Through our graphical understanding the commutativity rule also follows immediately, It doesn’t matter which of the 2 points come first when we draw a line that connects them, therefore P+Q = Q+P.

Understanding associativity and closure

Neither associativity nor closure are obvious from our graph.  We could check individual points to show that (P+Q) + R = P+(Q+R), but it would be harder to explain why this always held.  Equally whilst it’s clear that P+Q will always create a point on the curve it’s not obvious that this will be a rational point.

In fact we do have both associativity and closure for our group as we have the following algebraic definition for our addition operation:

For a given point:

On a given curve of the form:

The addition of 2 points is given by:

In the case of our curve:

If we take P = (1,12).  P + P will be given by:

We can check this result graphically.  If P and Q are the same point, then the line that passes through both P and Q has to be the tangent to the curve at that point.  Therefore we would have:

Here the tangent at A does indeed meet the curve again – at point C, which does reflect in y = x to give us the coordinates above.

We could also find this intersection point algebraically.  If we differentiate the original curve to find the gradient when x = 1 we can find the equation of the tangent when x=1 and then substitute this back into the equation of the curve to find the intersection point.  This would give us:

We would then reverse the x and y coordinates to reflect in the line y = x.  This also gives us the same coordinates.

More generally if we have the 2 rational coordinates on the curve:

We have the algebraic formula for addition as:


If P = (1,12) and Q = (9,10), P + Q would give (after much tedious substitution!):

This agrees with the coordinates we found earlier using the much easier geometrical approach. As we can see from this formula, both coordinate points will always be rational – as they will be composed of combinations of our original rational coordinates.  For any given curve there will be a generator set of coordinates through which we can generate all other rational coordinates on the curve through our addition operation.

So, we seem to have come a long way from our original goal – finding integer solutions to an algebraic equation. Instead  we seem to have got sidetracked into studying graphs and establishing groups.  However by reinterpreting this problem as one in group theory then this then opens up many new mathematical techniques to help us understand the solutions to this problem.

A fuller introduction to this topic is the very readable, “Taxicabs and the Sum of Two Cubes” by Joseph Silverman (from which the 2 general equations were taken) .

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