Sums of Squares Revisited

My first post for this mathematics blog concerned the conditions required for a number to be expressible as the sum of two squares. The date was 26th September 2015. I thought I might start revisiting some of my earlier posts just to reacquaint myself with their content. Let's consider prime numbers first. Fermat's theorem on the sums of two squares states that any prime number that is congruent to 1 modulus 4 can be expressed as a sum of two squares.

In other words, if \(p\) is a prime, then \(p=x^2+y^2\) where \(x\) and \(y\) are integers, if and only if \(p \equiv 1 \pmod{4}\). Applied to my current day count of \(24799\) (a prime), we find that \(24799 \equiv 3 \pmod{4}\) and so it cannot be written as the sum of two squares. In fact, it cannot even be written as the sum of three squares because the number can be expressed in the form \(4^a(8b+7)\) with \(a=0\) and \(b=3099\). By Legendre's 3-square theorem, such a number cannot be expressed as a sum of three squares.

Primes that can be expressed as the sum of two squares are called Pythagorean Primes.

If the number is composite, none of its \(4k+3\) primes can have an odd exponent. For example, two days ago I was \(24797\) days old and this number factorises to \(137 \times 181\). Now \(137 \equiv 1 \pmod{4}\) and \(181 \equiv 1 \pmod{4}\), so there are no \(4k+3\) primes and thus the number is expressible as a sum of two squares (in two different ways as it turns out): \(24797=59^2+146^2=74^2+139^2\).

There is a reason that \(24797\) can be expressed as a sum of squares in two different ways. Remember that its factors \(137\) and \(181\) are both primes satisfying \(p \equiv 1 \pmod{4}\) and so each can be expressed as a sum of two squares. Specifically, \(137=4^2+11^2\) and \(181=9^2+10^2\). It can easily be shown that the product of two sums of squares is equal to a sum of squares in two different ways. Here is the demonstration: $$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2=(ac+bd)^2+(ad-bc)^2$$There is an interesting history attached to this identity:

It is actually first found in Diophantus' Arithmetica (III, \(19\)), of the third century A.D. It was rediscovered by Brahmagupta (\(598\)–\(668\)), an Indian mathematician and astronomer, who generalised it (to the Brahmagupta identity) and used it in his study of what is now called Pell's equation. His Brahmasphutasiddhanta was translated from Sanskrit into Arabic by Mohammad al-Fazari, and was subsequently translated into Latin in \(1126\). The identity later appeared in Fibonacci's Book of Squares in \(1225\). Source.

However, take yesterday's number \(24798=2 \times 3 \times 4133\). Clearly, \(3\) is a \( 4k+3\) prime raised to an odd power and so \(24798\) cannot be expressed as sum of two squares. As it happens, \(4133 \equiv 1 \pmod{4}\) but that doesn't matter because \(3\) has already ruined things.

LaTeX and Integrating Square Root of Tangent Function

The following integration follows the steps outlined in this YouTube video that I watched, so I'm not claiming any originality here. However, I wanted to practise my LaTeX skills and so, using MacTex, a LaTeX implementation for OS X, and TeXShop, the front-end text editor, I translated the steps outlined in the video into LaTeX. MacTex produced a crisp PDF output file but later I wondered what would happen if I copied the code and pasted it into this blog. Well, turns out it rendered almost perfectly. The result is shown below:

**********************************************

How can we deal with  \( \! \! \int \sqrt{\tan x} \, \mathrm {d} x \)? If the square root sign wasn't there it would be easy and we'd get \( \ - \log(\cos x)+C \). However, it is there and we need to find a way to deal with it. Well, it turns out that there is a judicious substitution that we can use to get the process started and that substitution is \( u=\sqrt{\tan x} \).

This means that \( u^2=\tan x \) and so \( {2 \, u \, \mathrm {d} u} /{\mathrm {d}  x}=\sec^2 x \). \( \mathrm{d}x \)can now be written as \( 2 \, u \, \mathrm {d} u /sec^2 x \) and thus the integral becomes \( \! \! \int 2 \, u^2 \, \mathrm{d} u/\sec^2x \)and the problem now is to express \( \sec^2 x \) in terms of \(u\). Fortunately that's not too difficult with recourse to a little trigonometry. We know \( \tan x=u^2 \) and if we write this as \( \tan x=u^2/1 \) (opposite over adjacent) then the hypotenuse of the associated right-angled triangle is \( \sqrt{1+u^4} \).

Thus \( \cos x=1/\sqrt{1+u^4} \), \( \sec^2 x=1+u^4 \) and the integral becomes: $$\! \! \int \frac{2 \, u^2}{1+u^4} \mathrm {d} u$$There is still a way to go before we arrive at the solution to our problem but this is a far more manageable integral than the one that we started with. Now we need to spend some time manipulating the integral to get it into the right form. Firstly, let's divide top and bottom by \(u^2\). This gives us: $$\! \! \int \frac {2} {u^2+1/u^2} \mathrm {d} u$$The next step is to us modify the numerator 2 so that it looks like this:$$1+1/u^2+1-1/u^2$$Now we separate the integral into two halves as follows:$$\! \! \int \frac {1 + 1/u^2}{u^2+1/u^2} \mathrm {d} u \quad + \quad \! \! \int \frac {1 - 1/u^2}{u^2+1/u^2} \mathrm {d} u$$The denominators in both integrals can be rewritten as shown:$$\! \! \int \frac {1 + 1/u^2}{(u-1/u)^2+2} \mathrm {d} u \quad + \quad \! \! \int \frac {1 - 1/u^2}{(u+1/u)^2-2} \mathrm {d} u$$Each integral needs to be dealt with separately so let's begin with the integral on the left hand side. However, we need to make another substitution before we proceed. Let $t=u-1/u$. Differentiating both sides w.r.t. u, gives $\mathrm {d} t/\mathrm{d}u=1+1/u^2$ and so the integral on the left hand side can be rewritten:$$\! \! \int \frac {1 + 1/u^2}{(u-1/u)^2+2} \mathrm {d} u=\! \! \int \frac {1}{t^2+2} \mathrm {d} t$$An integral of the form $$\! \! \int \frac {1}{t^2+a^2} \mathrm {d} t =\frac{1}{a} \tan^{-1} \bigg (\frac{t}{a}\bigg ) + C $$and so we end up with $$\frac{1}{\sqrt{2}}\tan^{-1} \bigg (\frac {t}{\sqrt{2}}\bigg ) + C $$Now we need to backtrack, replacing $t$ by $u$ and $u$ by $\sqrt{\tan x}$. This will give us firstly $$\frac{1}{\sqrt{2}}\tan^{-1} \bigg (\frac {u-1/u}{\sqrt{2}}\bigg ) + C $$And then $$\frac{1}{\sqrt{2}}\tan^{-1} \bigg (\frac {\sqrt{\tan x}-1/\sqrt{\tan x}}{\sqrt{2}}\bigg ) + C $$But this is only the left hand side of our integral! Now we need to go back and deal with the right hand side integral which was:$$\quad \! \! \int \frac {1 - 1/u^2}{(u+1/u)^2-2} \mathrm {d} u$$This time let $t=u+1/u$. Differentiating both sides w.r.t. u, gives $\mathrm {d} t/\mathrm{d}u=1-1/u^2$ and so the integral on the left hand side can be rewritten:$$\quad \! \! \int \frac {1 - 1/u^2}{(u+1/u)^2-2} \mathrm {d} u=\! \! \int \frac {1}{t^2-2} \mathrm {d} t$$This integral can be solved by rewriting $t^2-2$ as $(t-\sqrt{2})(t+\sqrt{2})$ and then using the method of partial fractions to find A and B in the expression:$$\frac{A}{t-\sqrt{2}}+\frac{B}{t+\sqrt{2}}$$This leads to $(A+B)t+(A-B)\sqrt{2}=1$, $A=-B$, $2A\sqrt{2}=1$ and $A=1/2\sqrt{2}$

Hence $B=-1/2\sqrt{2}$ and the integral can be now be rewritten as:$$\int \frac{1/2\sqrt{2}}{t-\sqrt{2}} \mathrm {d} t-\int \frac{1/2\sqrt{2}}{t+\sqrt{2}} \mathrm {d} t$$Both these integrals are natural logarithms and so the result obtained is:

$$\frac{1}{2\sqrt{2}}(\ln (t-\sqrt{2})-\ln (t+\sqrt{2}))+C$$This simplifies to:$$\frac{1}{2\sqrt{2}}\ln \bigg(\frac {t-\sqrt{2}}{t+\sqrt{2}} \bigg )+C$$Now we need to backtrack, replacing $t$ by $u$ and $u$ by $\sqrt{\tan x}$. This will give us firstly$$\frac{1}{2\sqrt{2}}\ln \bigg(\frac {(u+1/u)-\sqrt{2}}{(u+1/u)+\sqrt{2}} \bigg )+C$$And then $$\frac{1}{2\sqrt{2}}\ln \bigg(\frac {(\sqrt{\tan x}+1/\sqrt{\tan x})-\sqrt{2}}{(\sqrt{\tan x}+1/\sqrt{\tan x})+\sqrt{2}} \bigg )+C$$Putting both integrals together now we get the result that:$$\! \! \int \sqrt{\tan x} \mathrm {d} x=\frac{1}{\sqrt{2}}\tan^{-1} \bigg (\frac {\sqrt{\tan x}-1/\sqrt{\tan x}}{\sqrt{2}}\bigg )+ \\ \frac{1}{2\sqrt{2}}\ln \bigg(\frac {(\sqrt{\tan x}+1/\sqrt{\tan x})-\sqrt{2}}{(\sqrt{\tan x}+1/\sqrt{\tan x})+\sqrt{2}} \bigg )+C$$

It was an arduous process but in the end our goal was accomplished.

General Solution to Cubic Equation

In a recent post titled All Cubic Polynomials Are Point Symmetric, the key point was that the \(x\) coordinate of the point of symmetry is given by: $$ x=\frac{-b}{3a}$$ and this fact also plays a crucial role in devising a solution to the general cubic equation \( ax^3+bx^2+cx+d=0\). It enables the point of symmetry in the graph of the polynomial to be translated horizontally so it lies on the \(y\) axis and the new equation is much easier to solve because the \(x^2\) term is removed.

Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving a cubic equation, on the other hand, was the first major success story of Renaissance mathematics in Italy. The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna. Source

The technique finds a single real root and the other two roots (real or complex) can be found by polynomial division and the use of the quadratic formula. The first step in the solution is to horizontally translate the point of symmetry so it lies on the \(y\) axis. This is achieved by means of the substitution:$$x=y-\frac{b}{3a}$$This produces:$$a(y-\frac{b}{3a})^3+b(y-\frac{b}{3a})^2+c(y-\frac{b}{3a})+d=0$$Multiplying this out gives$$ay^3+(c-\frac{b^2}{3a})y+(d+\frac{2b^2}{27a^2}-\frac{bc}{3a})=0$$After division by \(a\), this equation is of the general form \(y^3+Ay=B\) and was solved by Scipione del Ferro (1465-1526).

Consider a specific cubic equation e.g. \(x^3-15x^2+81x-175=0\).

Here the substitution will be \(x=y+5\) because \(a=1\) and \(b=-15\) leading to \(y^3+6y-20=0\). The situation is shown in the graph below, where it can be seen that the root of the original equation is 7 and that of the translated equation is 2 (although algebraically we don't know that yet):

We need to find \(s\) and \(t\) such that \(3st=A\) and \(s^3-t^3=B\).

It can be shown that \(y=s-t\) is then a solution of \(y^3+Ay=B\):$$(s-t)^3+3st(s-t)=s^3-t^3$$ $$ (s^3-3s^2t+3st^2-t^3)+(3s^2t-3st^2)=s^3-t^3$$Let's solve \(3st=A\) and \(s^3-t^3=B\) for the simplified cubic \(y^3+6y=20\).

\(3st=6\) and so \(s=2/t\).

Substituting into \(s^3-t^3=20\) gives \((2/t)^3-t^3=20\).

Multiplying by \(t^3\) gives \(t^6+20t^3-8=0\)

Taking the positive root \(t^3=-10+\sqrt{108}\) and \(t=\sqrt[3]{-10+\sqrt{108}}\)

Now \(s^3=20+t^3\) and so \(s^3=20+(-10+\sqrt{108})=10+\sqrt{108}\)

This means that \(s=\sqrt[3]{10+\sqrt{108}}\) and then, because \(y=s-t\), we have:

\(y=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}\)

But \(x=y+5\) and so \(x=\sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}+5\)

Remarkably, this formidable root (the only real root) evaluates to 7!

So that means \( \sqrt[3]{10+\sqrt{108}}-\sqrt[3]{-10+\sqrt{108}}=2\). Why is it so?

Well, here is how it all comes about:

\(\sqrt[3]{10 + \sqrt{108}} - \sqrt[3]{\sqrt{108} - 10}\)

\(=\sqrt[3]{10 + 6 \sqrt{3}} - \sqrt[3]{6 \sqrt{3} - 10}\)

\(=\sqrt[3]{1^3+3\sqrt{3}+3(\sqrt{3})^2+(\sqrt{3})^3}-\sqrt[3]{-1^3+3\sqrt{3}-3(\sqrt{3})^2+(\sqrt{3})^3}\)

\(=\sqrt[3]{(1+\sqrt{3})^3}-\sqrt[3]{(-1+\sqrt{3})^3}\)

\(=1+\sqrt{3}+1-\sqrt{3}\)

\(=2\)

Reference: The "Cubic Formula"

Number Bases (Radices)

In mathematical numeral systems, the radix or base is the number of unique digits, including zero, used to represent numbers in a positional numeral system. For example, for the decimal system (the most common system in use today) the radix is ten, because it uses the ten digits from 0 through 9. Source

Today I turned 24793 days old, a prime number of days, and as it turns out a palindrome in base 12 (12421). I started playing around with the representation of 24793 in other number bases or radices. In base 31, the representation is poo so I may be in for a shitty day. In base 36, the letters of the alphabet are exhausted:

Beyond base 36, the following system is used:

More generally, in a system with radix b (b > 1), a string of digits \(\ d_1 … d_n \) denotes the number \(\ d_1b^{n−1}+d_2b^{n−2}+ … +d_nb^{0} \), where \(\ 0 ≤ d_i < b \).

In practice, a colon is used to separate the individual digits e.g. 24793 is written\(\ 18:4:3 \,_{37} \)  in base 37 which can be dispensed with of course in the case of base 10.

Radices are usually natural numbers but they don't have to be. For example, a base using the golden ratio\(\ \Phi\) is possible, remembering that\(\ \Phi\) is given by the expression: \[\ \frac{1+\sqrt{5}}{2}\] Commonly, the capital letter\(\ \Phi\) is used to represent 1.618033988749895… and the lower case letter \( \phi\) to represent 0.618033988749895… or\(\ \Phi-1\) but this is certainly not always the case.

Such a base is colloquially called phinary and the following table gives some idea of how it works (source) but uses\(\ \varphi\), a variation of the lowercase\(\ \phi\) but meant to equal 1.618033988749895… here:

\(\ \Phi\) is closely linked to the Fibonacci sequence since \[ \lim_{n \to \infty} \frac{F_n}{F_{n-1}}=\Phi \]More information about \( \Phi  \) as a number base can be found on this site.

All Cubic Polynomials Are Point Symmetric

It was accident that I stumbled upon this PDF file discussing point symmetry in cubic polynomials. I had been looking for information about sketching cubic polynomials that could help a student I am currently tutoring in Mathematics. I've extracted the key observation in the screen shot below:

Here is an annotated graph that I produced in GeoGebra illustrating the point of symmetry for a specific cubic polynomial: \(\ f(x)=x^3-3x^2+2x-1\)

A translation of the graph (-1, 1) to (0, 0) makes the graph symmetry about the origin and an odd function. The new equation is calculated as follows: \(\ f(x)-1=(x+1)^3-3(x+1)^2+2(x+1)-1\) and \(\ f(x)=x(x+1)(x-1)\) or \(\ f(x)=x(x^2-1)\).

Harshad Numbers

Today I turned 24786 days old and one of the occurrences of this number in the OEIS was in A097569: right-truncatable Harshad numbers (zeros not permitted). Of course, I needed to ask myself the inevitable question: what is a Harshad number? Here is what Wikipedia had to say about it:

In recreational mathematics, a harshad number (or Niven number) in a given number base, is an integer that is divisible by the sum of its digits when written in that base. Harshad numbers in base \(\displaystyle n\) are also known as \(\displaystyle n\)-harshad (or \(\displaystyle n\)-Niven) numbers. Harshad numbers were defined by D. R. Kaprekar, a mathematician from India. The word "harshad" comes from the Sanskrit harṣa (joy) + da (give), meaning joy-giver. The term “Niven number” arose from a paper delivered by Ivan M. Niven at a conference on number theory in 1977. 

Stated mathematically, let \(\displaystyle X\) be a positive integer with \(\displaystyle m\) digits when written in base \(\displaystyle n\), and let the digits be \(\displaystyle a_{i}\) ( \(\displaystyle i\) = 0, 1, ..., \(\displaystyle m\) − 1). (It follows that \(\displaystyle a_{i}\) must be either zero or a positive integer up to \(\displaystyle n\) − 1.) \(\displaystyle X\) can be expressed as \[\displaystyle X=\sum _{i=0}^{m-1}a_{i}n^{i}\]
If there exists an integer \(\displaystyle A\) such that the following holds, then \(\displaystyle X\) is a harshad number in base \(\displaystyle n\): \[\displaystyle X=A\sum _{i=0}^{m-1}a_{i}\] A number which is a harshad number in every number base is called an all-harshad number, or an all-Niven number. There are only four all-harshad numbers: 1, 2, 4, and 6 (The number 12 is a harshad number in all bases except octal).

So, getting back to 24786, it can be seen that it is a Harshad number because 2+4+7+8+6 = 27 and 24786/27 = 918. Truncating the last digit on the right gives 2478, so that 2+4+7+8 = 21 and 2478/21 = 118. Truncating again gives 247, so that 2+4+7 = 13 and 247/13 = 19. Truncating again gives 24, so that 2+4=6 and 24/6=4. Finally, 2 itself is of course a Harshad number.

Furthermore, it is asserted in the OEIS entry that 24786 is the final such number in the sequence of right-truncatable Harshad numbers. The full sequence is then given by:

1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 18, 21, 24, 27, 36, 42, 45, 48, 54, 63, 72, 81, 84, 126, 216, 243, 247, 364, 423, 481, 486, 846, 2478, 8463, 24786

Look and Say Sequence

I came across the so-called "look-and-say sequence" popularised by the mathematician John Conway while watching this YouTube video:

There is a good explanation of the sequence on Wikipedia, part of which I've copied below:

In mathematics, the look-and-say sequence is the sequence of integers beginning as follows: 

1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, ... 

This is sequence A005150 in the OEIS.

To generate a member of the sequence from the previous member, read off the digits of the previous member, counting the number of digits in groups of the same digit. For example: 

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.
1211 is read off as "one 1, one 2, then two 1s" or 111221.
111221 is read off as "three 1s, two 2s, then one 1" or 312211.

What Conway found was that the ratio of the lengths of successive terms tended toward a constant value of \(\lambda\) = 1.303577269034... or to express it more precisely:
$$\lim_{n \to \infty} \frac{L_{n+1}}{L_n} = \lambda$$ where \(L_n\) denotes the number of digits in the n-th term of the sequence. Moreover, Conway showed that \(\lambda\) is an algebraic number and in fact the positive solution of a polynomial of degree 71 described in this Wikipedia article.

Of course, if allowing complex number solutions, there will be 71 solutions and when plotted on the Argand diagram they look like this (notice \(\lambda\) on the far right of the positive x-axis):

Android, LaTeX, Termial and Surreal Numbers

Donald Ervin Knuth
January 10, 1938 (age 79)
Milwaukee, Wisconsin, U.S.

It was with great surprise that I discovered my LaTeX-based equations in the two previous posts were not displaying using the Chrome browser on my Android phone. I checked using the default Internet browser and the result was the same. I checked on my iPad, using both Chrome and Safari, and all was well. On Ubuntu (under VirtualBox), using Firefox, and in Windows (under VirtualBox), using Microsoft Edge and Chrome, there was again no problem.

It was only Android, at least version 5 which I'm using on my Note 3. Maybe there is support for LaTeX in the 6 and 7 versions of Android. However, there's surprisingly little information about this problem and certainly no open admission anywhere that LaTeX isn't supported by Android.

This leads on to the term termial. I stumbled across it as I researched LaTeX because its creator, Donald Knuth, invented the term in his The Art of Computer Programming. It was devised as an analogy of the factorial, so as to illustrate the extension of the latter to the real numbers. It's defined as follows:

Let n \(\in\) ℤ > 0 (that is a positive integer) then n? = \(\sum_{k=1}^{n} k \)

Let n \(\in\) ℝ (that is a real number) then n? = \(\frac{n(n+1)}{2}\)

So 4? = 4 + 3 + 2 + 1 = 10 and \(\frac{1}{2}\)? = \(\frac{3}{8}\)

The second formula works for integers because 4? = 4 x 5 ÷ 2 = 10.

Donald Knuth is certainly an interesting character. This is a link to his biography on Wikipedia. He wrote a book called Surreal Numbers and below is a video in which he discusses the inspiration behind the idea of surreal numbers and how he actually set about writing the book in six days.

The book is freely available on the Internet and I've downloaded a copy and added it to my library.

MacTeX

If looking at this post on an Android device, the equations will not display properly.
A small but nagging problem has led me deeper into the arcana of LaTeX. The problem was that I wanted the following equation to display on the left and not in the centre:
\[x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
As I investigated further, frequent mention was made of the amsmath package as an add-on to LaTeX but this turned out to be only one of many additional add-ons. It was recommended that MacTeX be installed as this package contained all available packages rolled into one installation.

And so I found a MacTeX package at https://tug.org/mactex/mactex-download.html and promptly downloaded and installed it. The downloaded package is 2.8 GB and the installation takes up over 5 GB so it's a big affair.

The front-end to MacTex is TeXShop and it allows for full formatting of documents and produces PDF output. However, it does not seem suited for web page design which is my focus at the moment.

Using LaTeX in Blogger

$$x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$I've posted to my other, more educationally oriented blog Pedogogical Posturing about how to create mathematical expressions in Blogger but I've more to add on this mathematics specific blog. Above is the quadratic formula and below are the Greeks letters rendered using LaTeX:

• Alpha: A and \(\alpha \)
• Beta: B and \(\beta\)
• Gamma: \(\Gamma\) and \(\gamma\)
• Delta \(\Delta\) and \(\delta\)
• Epsilon: E and \(\epsilon\) and \(\varepsilon\)
• Zeta: Z and \(\zeta\)
• Eta: H and \(\eta\)
• Theta: \(\Theta\) and \(\theta\) and \(\vartheta\)
• Iota: I and \(\iota\)
• Kappa: K and \(\kappa\) and \(\varkappa\)
• Lambda: \(\Lambda\) and \(\lambda\)
• Mu: M and \(\mu\)
• Nu: N and \(\nu\)
• Xi: \(\Xi\) and \(\xi\)
• O: O and o
• Pi: \(\Pi\) and \(\pi\) and \(\varpi\)
• Rho: P and \(\rho\) and \(\varrho\)
• Sigma: \(\Sigma\) and \(\sigma\) and \(\varsigma\)
• Tau: T and \(\tau\)
• Upsilon: \(\Upsilon\) and \(\upsilon\)
• Phi: \(\Phi\) and \(\phi\) and \(\varphi\)
• Chi: X and \(\chi\)
• Psi: \(\Psi\) and \(\psi\)
• Omega: \(\Omega\) and \(\omega\)

The following fraction is nicely rendered as a continued fraction using LaTeX: \[\begin{equation} \frac{13}{8} = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{2} } } } \end{equation}\] I'll certainly be making more use of LaTeX in future posts. Of course, mathematical symbols can be generated using the UTF-8 character set, the default character encoding for HTML-5. Characters are generated using a decimal code prefixed by &# or a hexadecimal code prefixed by &#x e.g. &#8723 or &#x2213 should generate ∓ when inserted into HTML. However, this doesn't work in Blogger and instead one has to go to "Insert special characters" which is odd. 

A final observation is that the LaTeX is slow to be rendered when first opening the blog post. One can see the underlying code before the page is properly rendered. This is because the required javascript code is being fetched from:

http://cdn.mathjax.org/mathjax/latest/MathJax.js

Click on the link and you can see the code. It's also possible to download the code and inject it into your webpage or blog (if that's allowable), thus speeding up the rendering process.