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**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).

Essential resources for IB students:

Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.

There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.

Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

2) Exploration Guides and Paper 3 Resources

I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

**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.

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?

Essential resources for IB students:

Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.

There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.

Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

2) Exploration Guides and Paper 3 Resources

I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

**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:

For:

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!

Essential resources for IB students:

Revision Village has been put together to help IB students with topic revision both for during the course and for the end of Year 12 school exams and Year 13 final exams. I would strongly recommend students use this as a resource during the course (not just for final revision in Y13!) There are specific resources for HL and SL students for both Analysis and Applications.

There is a comprehensive Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and then provides a large bank of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial. Really useful!

The Practice Exams section takes you to a large number of ready made quizzes, exams and predicted papers. These all have worked solutions and allow you to focus on specific topics or start general revision. This also has some excellent challenging questions for those students aiming for 6s and 7s.

Each course also has a dedicated video tutorial section which provides 5-15 minute tutorial videos on every single syllabus part – handily sorted into topic categories.

2) Exploration Guides and Paper 3 Resources

I’ve put together four comprehensive pdf guides to help students prepare for their exploration coursework and Paper 3 investigations. The exploration guides talk through the marking criteria, common student mistakes, excellent ideas for explorations, technology advice, modeling methods and a variety of statistical techniques with detailed explanations. I’ve also made 17 full investigation questions which are also excellent starting points for explorations. The Exploration Guides can be downloaded here and the Paper 3 Questions can be downloaded here.

**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 a^{2}+b^{2} = c^{3}+d^{3} = e^{4}+f^{4}.

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:

a^{2}+b^{2} = c^{3}+d^{3} = e^{4}+f^{4}.

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 = 2^{6}, b = 2^{6}, which gives 2^{12}+2^{12}

c = 2^{4}, d = 2^{4}, which gives 2^{12}+2^{12}

e = 2^{3}, f = 2^{3}, which gives 2^{12}+2^{12}

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:

list1=[]

for a in range(2, 200):

for b in range(2,200):

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,e,f)

This returns the following solutions: 8192, 18737, 76832. Of these we reject the first as this is the solution 2^{12}+2^{12} 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 199^{2} + 199^{2}) is:

18737 = 64^{2}+121^{2} = 17^{3}+24^{3} = 11^{4}+8^{4}.

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 g^{5}+h^{5}?

Essential resources for IB students:

2) Exploration Guides and Paper 3 Resources

**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 a^{3}-b^{3}

**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 100^{2}-9^{2} or 22^{3}-9^{3} or 10^{4}-3^{4}.

14625: Which can be formed as either 121^{2}-4^{2} or 25^{3}-10^{3} or 11^{4}-2^{4}.

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:

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) .

**Stacking cannonballs – solving maths with code**

Numberphile have recently done a video looking at the maths behind stacking cannonballs – so in this post I’ll look at the code needed to solve this problem.

**Triangular based pyramid.**

A triangular based pyramid would have:

1 ball on the top layer

1 + 3 balls on the second layer

1 + 3 + 6 balls on the third layer

1 + 3 + 6 + 10 balls on the fourth layer.

Therefore a triangular based pyramid is based on the sum of the first n triangular numbers.

The formula for the triangular numbers is:

and the formula for the sum of the first n triangular numbers is:

We can simplify this by using the identity for the sum of the first n square numbers and also the identity for the sum of the first n natural numbers:

Therefore:

and the question we want to find out is whether there is triangular based pyramid with a certain number of cannonballs which can be rearranged into a triangular number i.e.:

here n and m can be any natural number. For example if we choose n = 3 and m = 4 we see that we have the following:

Therefore we can have a triangular pyramid of height 3, which has 10 cannonballs. There 10 cannonballs can then be rearranged into a triangular number.

**Square based pyramids and above.**

For a square based pyramid we would have:

1 ball on the top layer

1 + 4 balls on the second layer

1 + 4 + 9 balls on the third layer

1 + 4 + 9 + 16 balls on the fourth layer.

This is the sum of the first n square numbers. So the formula for the square numbers is:

and the sum of the first n square numbers is:

**For a pentagonal based pyramid we have:**

1 ball on the top layer

1 + 5 balls on the second layer

1 + 5 + 12 balls on the third layer

1 + 5 + 12 + 22 balls on the fourth layer.

This is the sum of the first n pentagonal numbers. So the formula for the pentagonal numbers is:

and the formula for the first n pentagonal numbers is:

**For a hexagonal based pyramid we have:**

The formula for the first n hexagonal numbers:

and the formula for the sum of the first n hexagonal numbers:

For a **k-agon based pyramid we have**

and the formula for the sum of the first n k-agon numbers:

Therefore the general case is to ask if a k-agonal pyramid can be rearranged into a k-agon number i.e:

**Computers to the rescue**

We can then use some coding to brute force some solutions by running through large numbers of integers and seeing if any values give a solution. Here is the Python code. Type it (taking care with the spacing) into a Python editor and you can run it yourself.

You can then change the k range to check larger k-agons and also change the range for a and b. Running this we can find the following. (The first number is the value of k, the second the height of a k-agonal pyramid, the third number the k-agon number and the last number the number of cannonballs used).

**Solutions:**

3 , 3 , 4 , 10

3 , 8 , 15 , 120

3 , 20 , 55 , 1540

3 , 34 , 119 , 7140

4 , 24 , 70 , 4900

6 , 11 , 22 , 946

8 , 10 , 19 , 1045

8 , 18 , 45 , 5985

10 , 5 , 7 , 175

11 , 25 , 73 , 23725

14 , 6 , 9 , 441

14 , 46 , 181 , 195661

17 , 73 , 361 , 975061

20 , 106 , 631 , 3578401

23 , 145 , 1009 , 10680265

26 , 190 , 1513 , 27453385

29 , 241 , 2161 , 63016921

30 , 17 , 41 , 23001

32 , 298 , 2971 , 132361021

35 , 361 , 3961 , 258815701

38 , 430 , 5149 , 477132085

41 , 204 , 1683 , 55202400

41 , 505 , 6553 , 837244045

43 , 33 , 110 , 245905

44 , 586 , 8191 , 1408778281

50 , 34 , 115 , 314755

88 , 15 , 34 , 48280

145, 162, 1191, 101337426

276, 26, 77, 801801)

322, 28, 86, 1169686

823, 113, 694, 197427385

2378, 103, 604, 432684460

31265, 259, 2407, 90525801730

For example we can see a graphical representation of this. When k is 6, we have a hexagonal pyramid with height 11 or the 22nd hexagonal number – both of which give a solution of 946. These are all the solutions I can find – can you find any others? Leave a comment below if you do find any others and I’ll add them to the list!

**Crack the Beale Papers and find a $65 Million buried treasure?**

The story of a priceless buried treasure of gold, silver and jewels (worth around $65 million in today’s money) began in January 1822. A stranger by the name of Thomas Beale walked into the Washington Hotel Virginia with a locked iron box, which he gave to the hotel owner, Robert Morriss. Morriss was to look after the box for Beale as he went off on his travels.

In May 1822 Morriss received a letter from Beale which stated that the box contained papers of huge value – but that they were encoded for protection. Beale went on to ask that Morriss continue to look after the box until his return. He added that if he did not return in the next 10 years then he had instructed a close friend to send the cipher key on June 1832. After that time Morriss would be able to decipher the code and learn of the box’s secrets.

Well, Beale never returned, nor did Morriss receive the promised cipher key. Eventually he decided to open the box. Inside were three sheets of paper written in code, and an explanatory note. The note detailed that Beale had, with a group of friends discovered a seam of gold and other precious metals in Santa Fe. They had mined this over a number of years – burying the treasure in a secret location for safe keeping. The note then explained that the coded messages would give the precise location of the treasure as well as detailing which men were due a share.

Morriss devoted many years to trying to decipher the code in vain – before deciding at the age of 84 in 1862 that he should share his secret with a close friend. That friend would later publish the Beale Papers in 1885. The pamphlet that was published stirred huge interest in America – inspiring treasure hunters and amateur cryptographers to try and crack the code. The second of the 3 coded messages was cracked by the author of the pamphlet using what is known as a book code. The United States Declaration of Independence was used as the book to encode the message above.

The first number 115 refers to the 115th word in the Declaration of Independence, which is the word “instituted”. Therefore the first letter of the decoded message is “I”. The second number is 73, which refers to the 73rd word in the declaration – which is “hold”, so the second letter of the decoded message is “h”. Following this method, the following message was revealed:

*I have deposited in the county of Bedford, about four miles from Buford’s, in an excavation or vault, six feet below the surface of the ground, the following articles, belonging jointly to the parties whose names are given in number three, herewith:*

*The first deposit consisted of ten hundred and fourteen pounds of gold, and thirty-eight hundred and twelve pounds of silver, deposited Nov. eighteen nineteen. The second was made Dec. eighteen twenty-one, and consisted of nineteen hundred and seven pounds of gold, and twelve hundred and eighty-eight of silver; also jewels, obtained in St. Louis in exchange for silver to save transportation, and valued at thirteen thousand dollars.*

*The above is securely packed in iron pots, with iron covers. The vault is roughly lined with stone, and the vessels rest on solid stone, and are covered with others. Paper number one describes the exact locality of the vault, so that no difficulty will be had in finding it. Source*

After the pamphlet was published there was great interest in cracking the 2 remaining papers, an interest which has persisted into modern times. One of the uncracked papers is shown below:

In 1983 2 amateur treasure hunters were jailed for trying to dig up graves in Bedford, sure that they were about to find the missing gold. In 1989 a professional treasure hunter called Mel Fisher secretly bought a large plot of land after believing that the treasure was buried underneath. However nothing was found. Up until now all efforts to crack the code above have ended in failure. Perhaps the pamphlet was a giant hoax? Or perhaps the treasure is still waiting to be found.

The town of Bedford still receives visitors from around the world, keen to try and crack this centuries old puzzle. You can hire metal detectors and go looking for it yourself. The map above from 1891 shows the 4 mile radius from Buford’s tavern which is thought to contain the treasure. Maybe one day Beale’s papers will finally be cracked.

For more information on this topic read Simon Singh’s excellent The Code Book – which has more details on this case and many other code breaking puzzles throughout history.

If you want to try your own codebreaking skills, head over to our Schoolcodebreaking site – to test your wits against students from schools around the world!

Essential resources for IB students:

2) Exploration Guides and Paper 3 Resources

**Projective Geometry**

Geometry is a discipline which has long been subject to mathematical fashions of the ages. In classical Greece, Euclid’s elements (Euclid pictured above) with their logical axiomatic base established the subject as the pinnacle on the “great mountain of Truth” that all other disciplines could but hope to scale. However the status of the subject fell greatly from such heights and by the late 18th century it was no longer a fashionable branch to study. The revival of interest in geometry was led by a group of French mathematicians at the start of the 1800s with their work on projective geometry. This then paved the way for the later development of non-Euclidean geometry and led to deep philosophical questions as to geometry’s links with reality and indeed just what exactly geometry was.

Projective geometry is the study of geometrical properties unchanged by projection. It strips away distinctions between conics, angles, distance and parallelism to create a geometry more fundamental than Euclidean geometry. For example the diagram below shows how an ellipse has been projected onto a circle. The ellipse and the circle are therefore projectively equivalent which means that projective results in the circle are also true in ellipses (and other conics).

Projective geometry can be understood in terms of rays of light emanating from a point. In the diagram above, the triangle IJK drawn on the glass screen would be projected to triangle LNO on the ground. This projection does not preserve either angles or side lengths – so the triangle on the ground will have different sized angles and sides to that on the screen. This may seem a little strange – after all we tend to think in terms of angles and sides in geometry, however in projective geometry distinctions about angles and lengths are stripped away (however something called the cross-ratio is still preserved).

We can see in the image above that a projection from the point E creates similar shapes when the 2 planes containing IJKL and ABCD are parallel. Therefore the Euclidean geometrical study of similar shapes can be thought of as a subset of plane positions in projective geometry.

Taking this idea further we can see that congruent shapes can be achieved if we have the centre of projection, E, “sent to infinity:” In projective geometry, parallel lines do indeed meet – at this point at infinity. Therefore with the point E sent to infinity we have a projection above yielding congruent shapes.

Projective geometry can be used with conics to associate every point (pole) with a line (polar), and vice versa. For example the point A had the associated red line, d. To find this we draw the 2 tangents from A to the conic. We then join the 2 points of intersection between B and C. This principle of duality allowed new theorems to be discovered simply by interchanging points and lines.

An example of both the symmetrical attractiveness and the mathematical potential for duality was first provided by Brianchon. In 1806 he used duality to discover the dual theorem of Pascal’s Theorem – simply by interchanging points and lines. Rarely can a mathematical discovery have been both so (mechanically) easy and yet so profoundly

beautiful.

**Brianchon’s Theorem**

**Pascal’s Theorem**

**Poncelet**

Poncelet was another French pioneer of projective geometry who used the idea of points and lines being “sent to infinity” to yield some remarkable results when used as a tool for mathematical proof.

**Another version of Pascal’s Theorem:**

Poncelet claimed he could prove Pascal’s theorem (shown above) where 6 points on a conic section joined to make a hexagon have a common line. He did this by sending the line GH to infinity. To understand this we can note that the previous point of intersection G of lines AB’ and A’B is now at infinity, which means that AB’ and A’B will now be parallel. This means that H being at infinity also creates the 2 parallel lines AC’. Poncelet now argued that because we could prove through geometrical means that B’C and BC’ were also parallel, that this was consistent with the line HI also being at infinity. Therefore by proving the specific case in a circle where line GHI has been sent to infinity he argued that we could prove using projective geometry the general case of Pascal’s theorem in any conic .

**Pascal’s Theorem with intersections at infinity:**

This branch of mathematics developed quickly in the early 1800s, sparking new interest in geometry and leading to a heated debate about whether geometry should retain its “pure” Euclidean roots of diagrammatic proof, or if it was best understood through algebra. The use of points and lines at infinity marked a shift away from geometry representing “reality” as understood from a Euclidean perspective, and by the late 1800s Beltrami, Poincare and others were able to incorporate the ideas of projective geometry and lines at infinity to provide their Euclidean models of non-Euclidean space. The development of projective geometry demonstrated how a small change of perspective could have profound consequences.

Essential resources for IB students:

2) Exploration Guides and Paper 3 Resources

This is a nice example of using some maths to solve a puzzle from the mindyourdecisions youtube channel (screencaptures from the video).

**How to Avoid The Troll: A Puzzle**

In these situations it’s best to look at the extreme case first so you get some idea of the problem. If you are feeling particularly pessimistic you could assume that the troll is always going to be there. Therefore you would head to the top of the barrier each time. This situation is represented below:

**The Pessimistic Solution:**

Another basic strategy would be the optimistic strategy. Basically head in a straight line hoping that the troll is not there. If it’s not, then the journey is only 2km. If it is then you have to make a lengthy detour. This situation is shown below:

**The Optimistic Solution:**

The expected value was worked out here by doing 0.5 x (2) + 0.5 x (2 + root 2) = 2.71.

The question is now, is there a better strategy than either of these? An obvious possibility is heading for the point halfway along where the barrier might be. This would make a triangle of base 1 and height 1/2. This has a hypotenuse of root (5/4). In the best case scenario we would then have a total distance of 2 x root (5/4). In the worst case scenario we would have a total distance of root(5/4) + 1/2 + root 2. We find the expected value by multiply both by 0.5 and adding. This gives 2.63 (2 dp). But can we do any better? Yes – by using some algebra and then optimising to find a minimum.

**The Optimisation Solution:**

To minimise this function, we need to differentiate and find when the gradient is equal to zero, or draw a graph and look for the minimum. Now, hopefully you can remember how to differentiate polynomials, so here I’ve used Wolfram Alpha to solve it for us. Wolfram Alpha is incredibly powerful -and also very easy to use. Here is what I entered:

and here is the output:

So, when we head for a point exactly 1/(2 root 2) up the potential barrier, we minimise the distance travelled to around 2.62 miles.

So, there we go, we have saved 0.21 miles from our most pessimistic model, and 0.01 miles from our best guess model of heading for the midpoint. Not a huge difference – but nevertheless we’ll save ourselves a few seconds!

This is a good example of how an exploration could progress – once you get to the end you could then look at changing the question slightly, perhaps the troll is only 1/3 of the distance across? Maybe the troll appears only 1/3 of the time? Could you even generalise the results for when the troll is y distance away or appears z percent of the time?

Essential resources for IB students:

2) Exploration Guides and Paper 3 Resources