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**

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:

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:

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

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).

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

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

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

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.

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

Substituting this back into the parametric equations gives:

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

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

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

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.

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)**

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!

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.

**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)t^{n}. Again for matters of slight convenience I’ll look at the similar relationship -0.5(n+1)(n+2)at^{n}.

This gives (following the same method as above:

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:

When n = 1:

When n =2:

When n = 10:

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:**

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.

]]>**Projectile Motion III: Varying gravity**

We can also do some interesting things with projectile motion if we vary the gravitational pull when we look at projectile motion. The following graphs are all plotted in parametric form.

Here t is the parameter, v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which on Earth we will take as 9.81 m/s^{2}.

**Earth **

Say we take a projectile and launch it with a velocity of 10 m/s. When we vary the angle of launch we get the folowing graphs:

On the y axis we have the vertical height, and on the x axis the horizontal distance. Therefore we can see that the maximum height that we achieve is around 5m and the maximum horizontal distance is around 10m.

**Other planets and universal objects**

We have the following values for the gravitational pull of various objects:

Enceladus (Moon of Saturn): 0.111 m/s^{2}, The Moon: 1.62 m/s^{2}, Jupiter: 24.8 m/s^{2}, The Sun: 274 m/s^{2}, White dwarf black hole surface gravity: 7×10^{12}m/s^{2}.

So for each one let’s see what would happen if we launched a projectile with a velocity of 10 m/s. Note that the mass of the projectile is not relevant (though it would require more force to achieve the required velocity).

**Enceladus:**

**The Moon:**

**Jupiter:**

**The Sun:**

**Black hole surface gravity:**

This causes some issues graphically! I’ll use the equations derived in the last post to find the coordinates of the maximum point for a given launch angle theta:

Here we have v = 10 and g = 7×10^{12}m/s^{2}. For example if we take our launch angle (theta) as 45 degrees this will give the coordinates of the maximum point at:

(7.14×10^{-12}, 3.57×10^{-12}).

**Summary:**

We can see how dramatically life would be on each surface! Whilst on Earth you may be able to throw to a height of around 5m with a launch velocity of 10 m/s., Enceladus would see you achieve an incredible 450m. If you were on the surface of the Sun then probably the least of your worries would be how hight to throw an object, nevertheless you’d be struggling to throw it 20cm high. And as for the gravity at the surface of a black hole you wouldn’t get anywhere close to throwing it a nanometer high (1 billionth of a meter).

]]>

**Projectile Motion Investigation II**

Another example for investigating projectile motion has been provided by fellow IB teacher Ferenc Beleznay. Here we fix the velocity and then vary the angle, then to plot the maximum points of the parabolas. He has created a Geogebra app to show this (shown above). The locus of these maximum points then form an ellipse.

We can see that the maximum points of the projectiles all go through the dotted elliptical line. So let’s see if we can derive this equation.

Let’s start with the equations for projectile motion, usually given in parametric form:

Here v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which we will take as 9.81.

We can plot these curves parametrically, and for each given value of theta (the angle of launch) we will create a projectile motion graph. If we plot lots of these graphs for different thetas together we get something like this:

We now want to see if the maximum points are in any sort of pattern. In order to find the maximum point we want to find when the gradient of dy/dx is 0. It’s going to be easier to keep things in parametric form, and use partial differentiation. We have:

Therefore we find the partial differentiation of both x and y with respect to t. (This simply means we can pretend theta is a constant).

We can then say that:

We then find when this has a gradient of 0:

We can then substitute this value of t back into the original parametric equations for x:

and also for y:

We now have the parametric equations in terms of theta for the locus of points of the maximum points. For example, with g= 9.81 and v =1 we have the following parametric equations:

This generates an ellipse (dotted line), which shows the maximum points generated by the parametric equations below (as we vary theta):

And here is the graph:

We can vary the velocity to create a new ellipse. For example the ellipse generated when v =4 creates the following graph:

So, there we go, we have shown that different ellipses will be created by different velocities. If you feel like a challenge, see if you can algebraically manipulate the parametric equations for the ellipse into the Cartesian form!

]]>**Envelope of projectile motion**

For any given launch angle and for a fixed initial velocity we will get projectile motion. In the graph above I have changed the launch angle to generate different quadratics. The black dotted line is then called the envelope of all these lines, and is the boundary line formed when I plot quadratics for every possible angle between 0 and pi.

**Finding the equation of an envelope for projectile motion **

Let’s start with the equations for projectile motion, usually given in parametric form:

Here v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which we will take as 9.81.

First let’s rearrange these equations to eliminate the parameter t.

Next, we use the fact that the envelope of a curve is given by the points which satisfy the following 2 equations:

F(x,y,theta)=0 simply means we have rearranged an equation so that we have 3 variables on one side and have made this equal to 0. The second of these equations means the partial derivative of F with respect to theta. This means that we differentiate as usual with regards to theta, but treat x and y like constants.

Therefore we can rearrange our equation for y to give:

and in order to help find the partial differential of F we can write:

We can then rearrange this to get x in terms of theta:

We can then substitute this into the equation for F(x,y,theta)=0 to eliminate theta:

We then have the difficulty of simplifying the second denominator, but luckily we have a trig equation to help:

Therefore we can simplify as follows:

and so:

And we have our equation for the envelope of projectile motion! As we can see it is itself a quadratic equation. Let’s look at some of the envelopes it will create. For example, if I launch a projectile with a velocity of 1, and taking g = 9.81, I get the following equation:

This is the envelope of projectile motion when I take the following projectiles in parametric form and vary theta from 0 to pi:

This gives the following graph:

If I was to take an initial velocity of 2 then I would have the following:

And an initial velocity of 4 would generate the following graph:

So, there we have it, we can now create the equation of the envelope of curves created by projectile motion for any given initial velocity!

**Other ideas for projectile motion**

There are lots of other things we can investigate with projectile motion. One example provided by fellow IB teacher Ferenc Beleznay is to fix the velocity and then vary the angle, then to plot the maximum points of the parabolas. He has created a Geogebra app to show this:

You can then find that the maximum points of the parabolas lie on an ellipse (as shown below).

See if you can find the equation of this ellipse!

]]>**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:

- Find the hypotenuse of triangle AGC.
- Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
- Therefore triangle AGC and GBD are similar, so length BG = 4. We can now use Pythagoras to find length BD.
- We can find length CD by Pythagoras.
- Now we have 3 sides of a triangle, CDB. This allows use to find angle BDC using the cosine rule.
- Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
- Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
- 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.
- 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?

]]>This post carries on from the previous post on the Mordell Equation – so make sure you read that one first – otherwise this may not make much sense. The man pictured above (cite: Wikipedia) is Louis Mordell who studied the equations we are looking at today (and which now bear his name).

In the previous post I looked at solutions to the difference between a cube and a square giving an answer of 2. This time I’ll try to generalise to the difference between a cube and a square giving an answer of k. I’ll start with the same method as from the previous post:

In the last 2 lines we outline the 2 possibilities, either b = 1 or b = -1. First let’s see what happens when b = 1:

This will only provide an integer solution for a if we have:

Which generates the following first few values for k when we run through m = 1, 2,3..:

k = 2, 11, 26, 47

We follow the same method for b = -1 and get the following:

Which generates the following first few values for k when we run through m = 1, 2,3…:

k = 4, 13, 28, 49

These are the values of k which we will be able to generate solutions to. Following the same method as in the previous post this generates the following solutions:

Let’s illustrate one of these results graphically. If we take the solutions for k = 13, which are (17,70) and (17,-70), these points should be on the curve x cubed – y squared = 13.

This is indeed the case. This graph also demonstrates how all solutions to these curves will have symmetrical solutions (e, f) and (e, -f).

We can run a quick computer program to show that this method does not find all the solutions for the given values of k, but it does ensure solutions will be found for the k values in these lists.

In the code solutions above, results are listed k, x, y, x cubed, y squared. We can see for example that in the case of k = 11 our method did not find the solution x = 3 and y = 4 (though we found x = 15 and y = 58). So, using this method we now have a way of finding *some* solutions for *some* values of k – we’ve not cracked the general case, but we have at least made a start!

**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.

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?

]]>

**Using Maths to model the spread of Coronavirus (COVID-19)**

This coronavirus is the latest virus to warrant global fears over a disease pandemic. Throughout history we have seen pandemic diseases such as the Black Death in Middle Ages Europe and the Spanish Flu at the beginning of the 20th century. More recently we have seen HIV responsible for millions of deaths. In the last few years there have been scares over bird flu and SARS – yet neither fully developed into a major global health problem. So, how contagious is COVID-19, and how can we use mathematics to predict its spread?

Modelling disease outbreaks with real accuracy is an incredibly important job for mathematicians and all countries employ medical statisticians for this job . Understanding how diseases spread and how fast they can spread through populations is essential to developing effective medical strategies to minimise deaths. If you want to save lives maybe you should become a mathematician rather than a doctor!

Currently scientists know relatively little about the new virus – but they do know that it’s the same coronavirus family as SARS and MERS which can both cause serious respiratory problems. Scientists are particularly interested in trying to discover how infectious the virus is, how long a person remains contagious, and whether people can be contagious before they show any symptoms.

**In the case of COVID-19 we have the following early estimated values: **[From a paper published by medical statisticians in the UK on January 24]

**R _{0}. between 3.6 and 4.** This is defined as how many people an infectious person will pass on their infection to in a totally susceptible population. The higher the R

**Total number infected** by January 21: prediction interval 9,217–14,245. Of these an estimated 3,050–4,017 currently with the virus and the others recovered (or died). This is based on an estimation that only around 5% of cases have been diagnosed. By February 4th they predict 132,751–273,649 will be infected.

**Transmission rate β** estimated at 1.07. β represents the transmission rate per day – so on average an infected person will infect another 1.07 people a day.

**Infectious period** estimated at 3.6 days. We can therefore calculate μ (the per capita recovery rate) by μ = 1/(3.6). This tells us how quickly people will be removed from the population (either recovered and become immune or died)

**SIR Model**

The basic model is based on the SIR model. The SIR model looks at how much of the population is susceptible to infection (S), how many of these go on to become infectious (I), and how many of these are removed (R) from the population being considered (i.e they either recover and thus won’t catch the virus again, or die).

The Guardian datablog have an excellent graphic to show the contagiousness relative to deadliness of different diseases [click to enlarge, or follow the link]. We can see that seasonal flu has an R_{0}. value of around 2.8 and a fatality rate of around 0.1%, whereas measles has an R_{0}. value of around 15 and a fatality rate of around 0.3%. This means that measles is much more contagious than seasonal flu.

You can notice that we have nothing in the top right hand corner (very deadly and very contagious). This is just as well as that could be enough to seriously dent the human population. Most diseases we worry about fall into 2 categories – contagious and not very deadly or not very contagious and deadly.

The equations above represent a SIR (susceptible, infectious, removed) model which can be used to model the spread of diseases like flu.

dS/dt represents the rate of change of those who are susceptible to the illness with respect to time. dI/dt represents the rate of change of those who are infected with respect to time. dR/dt represents the rate of change of those who have been removed with respect to time (either recovered or died).

For example, if dI/dt is high then the number of people becoming infected is rapidly increasing. When dI/dt is zero then there is no change in the numbers of people becoming infected (number of infections remain steady). When dI/dt is negative then the numbers of people becoming infected is decreasing.

**Modelling for COVID-19**

N is the total population. Let’s take as the population of Wuhan as 11 million.

μ is the per capita recovery (Calculated by μ = 1/(duration of illness) ). We have μ = 1/3.6 = 5/18.

β the transmission rate as approximately 1.07

Therefore our 3 equations for rates of change become:

dS/dt = -1.07 S I /11,000,000

dI/dt = 1.07 S I /11,000,000 – 5/18 I

dR/dt = 5/18 I

Unfortunately these equations are very difficult to solve – but luckily we can use a computer program or spreadsheet to plot what happens. We need to assign starting values for S, I and R – the numbers of people susceptible, infectious and removed. With the following values for January 21: S = 11,000,000, I = 3500, R = 8200, β = 1.07, μ = 5/18, I designed the following Excel spreadsheet (instructions on what formula to use here):

This gives a prediction that around 3.9 million people infected within 2 weeks! We can see that the SIR model that we have used is quite simplistic (and significantly different to the expert prediction of around 200,000 infected).

So, we can try and make things more realistic by adding some real life considerations. The current value of β (the transmission rate) is 1.07, i.e an infected person will infect another 1.07 people each day. We can significantly reduce this if we expect that infected people are quarantined effectively so that they do not interact with other members of the public, and indeed if people who are not sick avoid going outside. So, if we take β as (say) 0.6 instead we get the following table:

Here we can see that this change to β has had a dramatic effect to our model. Now we are predicting around 129,000 infected after 14 days – which is much more in line with the estimate in the paper above.

As we are seeing exponential growth in the spread, small changes to the parameters will have very large effects. There are more sophisticated SIR models which can then be used to better understand the spread of a disease. Nevertheless we can see clearly from the spreadsheet the interplay between susceptible, infected and recovered which is the foundation for understanding the spread of viruses like COVID-19.

[Edited in March to use the newly designated name COVID-19]

]]>**Square Triangular Numbers**

Square triangular numbers are numbers which are both square numbers and also triangular numbers – i.e they can be arranged in a square or a triangle. The picture above (source: wikipedia) shows that 36 is both a square number and also a triangular number. The question is how many other square triangular numbers we can find?

The equation we are trying to solve is:

a^{2} = 0.5(b^{2}+b)

for some a, b as positive integers. The LHS is the formula to generate square numbers and the RHS is the formula to generate the triangular numbers.

We can start with some simple Python code (which you can run here):

```
```for c in range(1,10001):

for d in range(1,10001):

if c**2 == (d**2+d)/2:

print(c**2, c,d)

This checks the first 10000 square numbers and the first 10000 triangular numbers and returns the following:

1 1 1

36 6 8

1225 35 49

41616 204 288

1413721 1189 1681

48024900 6930 9800

i.e 1225 is the next square triangular number after 36, and can be formed as 35^{2} or as 0.5(49^{2}+49). We can see that there are very few square triangular numbers to be found in the first 50 million numbers. The largest we found was 48,024,900 which is made by 6930^{2} or as 0.5(9800^{2}+9800).

We can notice that the ratio between each consecutive pair of square triangular numbers looks like it converges as it gives:

36/1 = 36

1225/36 = 34.027778

41616/1225 = 33.972245

1413721/41616 = 33.970612

48024900/1413721 = 33.970564

So, let’s use this to predict that the next square triangular number will be around

48024900 x 33.9706 = 1,631,434,668.

If we square root this answer we get approximately 40391

If we solve 0.5(b^{2}+b) = 1,631,434,668 using Wolfram we get approximately 57120.

Therefore let’s amend our code to look in this region:

```
```for c in range(40380,40400):

for d in range(57100,57130):

if c**2 == (d**2+d)/2:

print(c**2, c,d)

This very quickly finds the next solution as:

1631432881 40391 57121

This is indeed 40391^{2} – so our approximation was very accurate. We can see that this also gives a ratio of 1631432881/48024900 = 33.97056279 which we can then use to predict that the next term will be 33.970563 x 1631432881 = 55,420,693,460. Square rooting this gives a prediction that we will use the 235,416 square number. 235,416^{2} gives 55,420,693,056 (using Wolfram Alpha) and this is indeed the next square triangular number.

So, using a mixture of computer code and some pattern exploration we have found a method for finding the next square triangular numbers. Clearly we will quickly get some very large numbers – but as long as we have the computational power, this method should continue to work.

**Using number theory**

The ever industrious Euler actually found a formula for square triangular numbers in 1778 – a very long time before computers and calculators, so let’s have a look at his method:

We start with the initial problem, and our initial goal is to rearrange it into the following form:

Next we make a substitution:

Here, when we get to the equation 1 = x^{2} – 2y^{2} we have arrived at a Pell Equation (hence the rearrangement to get to this point). This particular Pell Equation has the solution quoted above where we can define P_{k} as

Therefore we have

Therefore for any given k we can find the kth square triangular number. The a value will give us the square number required and the b value will give us the triangular number required. For example with k = 3:

This tells us the 3rd square triangular number is the 35th square number or the 49th triangular number. Both these give us an answer of 1225 – which checking back from our table is the correct answer.

So, we have arrived at 2 possible methods for finding the square triangular numbers – one using modern computational power, and one using the skills of 18th century number theory.

]]>