Prof R C Gupta and early Indian Mathematics

Yesterday I made an entry about the award of a prize to Prof R C Gupta for his huge opus of work in early Indian mathematics. I would like to say a bit more about this.

Prof Gupta has worked extensively on the work in mathematics and astronomy of the early mathematicians and astronomers like Aryabhata, Bhaskara I, Brahmagupta, Mahavira, Govindaswami, Bhaskara II and so on; also on early Jain mathematics in general. If you study the texts associated with all these mathematicians you do not see the subject developed as it is in today's books. Typically you see either extensive tables, or rules of some kind which are clearly the verbal equivalent of formulas, and these are generally given in the form of rhyming verse (in Sanskrit, of course). 

The first problem, obviously, is to interpret these verses correctly and meaningfully. But the really interesting question is: How did these mathematicians construct the formulas or the tables? 

For example, in Aryabhata's book Aryabhatiya, we find a table of first differences of the sine table, captured very compactly and conveniently in just two lines, in a form of verse. It seems amazing that in that distant era (5th century AD), Aryabhata was examining tables of differences of a function! And how did he even construct such a table? It compares very well with the modern table. Unfortunately, this question may never be fully answered, because the records are so scanty, but historians like Prof Gupta have looked very carefully at the question. (There are now many historians in the west too who have a great interest in such questions.) In the case of the later mathematicians (Bhaskara I, Mahavira, Bhaskara II) the records yield a little more. For example, it appears that some of these later mathematicians used second order interpolation methods to construct more accurate tables. Here is a paper of Prof Gupta's on just this topic, which you should have a look at: R C Gupta - 2nd order interpolation.

With regard to the Kerala school of mathematics - Madhava, Nilakantha and so on - the records are much clearer; in this school, the writers developed a tradition of describing how they got their results, in sharp contrast to (say) what Aryabhata or Bhaskara I did. So they describe clearly how they got the power series expansion for $\tan^{-1} x$. The famous series

\[ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \cdots \]

which is nowadays called the Leibnitz-Gregory series for $\pi$ should more correctly be called the Madhava series, or the Madhava-Gregory-Leibnitz series, because it appears in the work of the Kerala school - the Yuktibhasa - and this work predates the work of Gregory and Leibnitz by over two centuries.

The proof of the series for $\tan^{-1} x$ shows quite clearly that the Kerala school had an understanding of the limit process. What is not clear is whether they were able to generalize this notion and apply it to the study of functions in general. If they had done this, then they would have had the differential calculus in their hands, with all the power and glory that it brings. But it is probable that they did not; so in that sense one can say that these mathematicians "almost" discovered the calculus for themselves, but they did not quite get there.

A divisibility problem

It is embarrassing to see how for long I have not posted anything on this blog! But the summer break broke my momentum, and too many events intervened subsequently. Thankfully I am able to make an entry today, after a break of close to ten weeks.

A reader - Mr Vinod Kumar, lecturer in mathematics, of Payyanur College, Kannur District, Kerala - has sent in the following problem:


Consider all four digit numbers with the property that the sum of the first two digits is equal to the sum of the last two digits. Examples of such numbers are $4123$, $6372$ and $0413$. (In this problem we shall include the numbers with one, two or three digits among the four digit numbers - we simply consider their leading digits to be $0$. So we shall write $311$ as $0311$, $12$ as $0012$, and so on.)


Here is the problem: Show that the sum of all such four digit numbers is divisible by $11$.


A second question is: If we consider instead all the six digit numbers with the corresponding property - that is, the sum of the first three digits equals the sum of the last three digits - what would be the number that divides the sum of all such numbers? Once again, we pad the number with enough zeros from the left side to ensure that it has six digits.


Readers are invited to solve these problems!

Prof R C Gupta to be honored

The renowned math historian Prof R C Gupta was to be honored last year at the ICHM (Budapest, July 2009),

ICHM Special Session Henk Bos Valedictory Symposium

but he was unable to be present for the occasion. So the prize - the Kenneth O. May Prize - will be given to him this year instead - at the ICM to be held in Hyderabad (19-27 August 2010). Here is a link to the ICM:

http://www.icm2010.in/about-icm-2010

Prof Gupta has done an amazing amount of work on early Indian mathematics, and is a highly accomplished historian of mathematics. I have had the pleasure and honor of interacting with him (in Warangal in December 2007, during the annual AMTI conference), and can testify to how simple and unassuming a person he is.

A curious occurrence of the Fibonacci numbers

Anyone who has some familiarity with the Fibonacci numbers,
\[0, \,\ 1, \,\ 1, \,\ 2, \,\ 3, \,\ 5, \,\ 8, \,\ 13, \,\ 21, \,\ 34, \,\ 55, \,\ 89, \,\ \ldots,\]
will know that they have a way of cropping up in the most unlikely places. In today's posting we document another such instance.

They arise when we study the compositions of positive integers. I prefer to call them "ordered partitions" but "compositions" is the accepted term.

As you know, a partition of a positive integer $n$ is a way of writing $n$ as a sum of (an unspecified number of) positive integers, order not being regarded as important. Thus, there are $2$ partitions of $2$ (namely, $2$ and $1+1$); $3$ partitions of $3$:
\[3, \quad 2+1, \quad 1+1+1\]
and $5$ partitions of $4$:
\[4, \quad 3+1, \quad 2+2, \quad 2+1+1 \quad 1+1+1+1.\]
Note that we do not distinguish between $3+1$ and $1+3$; that is, the sums are unordered. The above lists show that the partition numbers of $2$, $3$, $4$ are $2$, $3$, $5$ respectively; we write $P(2) = 2$, $P(3) = 3$ , $P(4) = 5$, where $P$ denotes the partition function. The $P$-function grows at terrific speed; for example,

• $P(100) = 190,569,292$;
• $P(200) = 3,972,999,029,388$;
• $P(300) = 9,253,082,936,723,602$.

Big numbers!

The partition function is the setting for many famous and beautiful (and almost unbelievable) results, proved by mathematicians like Ramanujan, Hardy, Littlewood and Rademacher. But we shall not dwell on these results in this post.

A variation - "ordered partitions"

Instead, we shall choose to regard order as important, in which case we get ordered partitions (or compositions, as noted above). Let $f(n)$ denote the number of ordered partitions of $n$. We get the following values of $f(n)$:

• $f(1) = 1$, as $1$ has just one ordered partition ($1$ itself);
• $f(2) = 2$, as $2$ has the following two ordered partitions: $2$ and $1+1$;
• $f(3) = 4$, as $4$ has the following four ordered partitions: $3$, $2+1$,$1+2$, $1+1+1$;
• $f(4) = 8$, as $4$ has the following eight ordered partitions: $4$, $3+1$, $1+3$, $2+2$, $2+1+1$, $1+2+1$, $1+1+2$, $1+1+1+1$.

Observe the sequence of values: $1$, $2$, $4$, $8$. Why, we seem to have a doubling sequence! Can we expect, then, that $f(5) = 16$? Let's check: the ordered partitions of $5$ are $5$, $4+1$, $1+4$, $3+2$, $2+3$, $3+1+1$, $1+3+1$, $1+1+3$, $2+2+1$, $2+1+2$,$1+2+2$, $2+1+1+1$, $1+2+1+1$, $1+1+2+1$, $1+1+1+2$, $1+1+1+1+1$. The count is indeed $16$.

It appears then that $f(n) = 2^{n-1}$ for all $n \ge 1$. But how may one prove this compact and elegant result? Try it out!

Another variation - we leave out the number 1

Now let us introduce a simple variation: in the partitions we shall not allow the number $1$; thus, all the summands must exceed $1$. In this case $5$ has only three such ordered partitions: $5$, $3+2$, $2+3$.

In general, let $g(n)$ denote the number of ordered partitions of $n$ in which every summand exceeds $1$. We get the following values of $g(n)$:

• $g(1) = 0$ (no partitions are possible if $1$ cannot be used);
• $g(2) = 1$ (there is just one such partition, $2$ itself);
• $g(3) = 1$ (there is just one such partition, $3$ itself);
• $g(4) = 2$ (the two partitions are $4$ and $2+2$);
• $g(5) = 3$ (obtained above);
• $g(6) = 5$ (the five partitions are $6$, $4+2$, $2+4$, $3+3$, $2+2+2$);
• $g(7) = 8$ (the eight partitions are $7$, $5+2$, $2+5$, $4+3$, $3+4$, $3+2+2$, $2+3+2$, $2+2+3$);
• $g(8) = 13$ (the thirteen partitions are $8$, $6+2$, $2+6$, $5+3$, $3+5$, $4+4$, $4+2+2$, $2+4+2$, $2+2+4$, $3+3+2$, $3+2+3$, $2+3+3$, $2+2+2+2$).

We get the following sequence of values of the $g$ function: $0$, $1$, $1$, $2$, $3$, $5$, $8$, $13$. Why, we seem to have obtained the Fibonacci sequence! Is this really so?

Let us verify whether $g(9) = 21$ (for the next Fibonacci number after $13$ is $21$); the ordered partitions of $9$ in which every summand exceeds $1$ are $9$, $7+2$, $2+7$, $6+3$, $3+6$, $5+4$, $4+5$, $5+2+2$, $2+5+2$, $2+2+5$, $4+3+2$, $4+2+3$, $3+4+2$, $3+2+4$, $2+3+4$, $2+4+3$, $3+3+3$, $3+2+2+2$, $2+3+2+2$, $2+2+3+2$, $2+2+2+3$. Indeed there are $21$ such partitions!

Is it not remarkable that the Fibonacci numbers have arisen in such a context? To check that we are not being fooled by "circumstantial evidence", we need to prove logically that $g(n)$ is a Fibonacci number; the proof cannot be numerical or circumstantial. This is not as difficult as may seem at first sight, and may be shown by proving that the $g$'s obey the basic recursive law governing the Fibonacci numbers, namely:
\[g(n) = g(n-1) + g(n-2)\]
for any integer $n \ge 3$. Since $g(1)$ and $g(2)$ are already known to be the first two consecutive Fibonacci numbers, the rest then follows.

Here are some references to Fibonacci numbers that point out some further aspects of the Fibonacci numbers.

• Wolfram Mathworld - Fibonacci 
• Wikipedia - Mathworld 

The Wolfram reference lists several contexts where the Fibonacci numbers arise in a natural yet unexpected manner.

Pretty decimals

Everyone is fond of recurring decimals (well, I certainly am ...). Who can resist being charmed by the fact that
\[\frac{1}{7} = 0.142857 \ 142857 \ 142857 \ \ldots,\]
with the portion $142857$ recurring indefinitely (it is called the repetend) and having these several striking properties: first,

• $142857 \times 2 = 285714$
• $142857 \times 3 = 428571$
• $142857 \times 4 = 571428$
• $142857 \times 5 = 714285$
• $142857 \times 6 = 857142$

with the circular arrangement of digits in each product being exactly the same as in $142857$; and next, if we "slice" the number in two halves, like this:
\[\fbox{$142$} \ \fbox{$857$}\]
and add the two "halves", we get
\[142 + 857 = 999,\]
a number wholly composed of $9$'s.

If we try this with the fraction $1/13$, we find something very similar. The repetend is now $076923$, since
\[\frac{1}{13} = 0.076923 \ 076923 \ 076923 \ \ldots,\]
and this number has nearly the same features as the number $142857$; but there are also some curious differences. Thus, its multiples with the numbers $1$, $3$, $4$, $9$, $10$, $12$ yield numbers that have the same circular pattern of digits as $076923$:

• $076923 \times 1 = 076923$
• $076923 \times 3 = 230769$
• $076923 \times 4 = 307692$
• $076923 \times 9 = 692307$
• $076923 \times 10 = 769230$
• $076923 \times 12 = 923076$

whereas its multiples with the numbers $2$, $5$, $6$, $7$, $8$, $11$ yield another set of numbers which too bear the very same relationship amongst themselves:

• $076923 \times 2 = 153846$
• $076923 \times 5 = 384615$
• $076923 \times 6 = 461538$
• $076923 \times 7 = 538461$
• $076923 \times 8 = 615384$
• $076923 \times 11 = 846153$

Note that the numbers
\[153846, \quad 384615, \quad 461538, \quad 538461, \quad 615384, \quad 846153\]
all have the same circular arrangement of digits. How very curious!

You may well ask, is there any other property that bonds the numbers $1$, $3$, $4$, $9$, $10$, $12$ together, and similarly bonds the numbers $2$, $5$, $6$, $7$, $8$, $11$ together? Yes, there is! - try to find it.

Finally, note that if we slice the number $076923$ in two halves,
\[\fbox{$076$} \ \fbox{$923$}\]
and add the two halves, we get
\[076 + 923 = 999,\]
exactly like earlier.

Is this a general feature that we find with all recurring decimals? Try it out - go and explore more such decimals! You will almost certainly unearth more patterns of interest.

Here is an associated problem which is of some interest: Find the smallest positive integer with the feature that if the units digit is taken to the "opposite end" of the number, the new number is exactly twice the original number.

Powers sequences strung on a line

Infinite decimals have still more morsels to offer. For example, consider the decimal arising from the fraction $1/99998$:
\[\frac{1}{99998} = 0.00001 \  00002 \ 00004 \ 00008 \ 00016 \ 00032 \ 00064 \ \ldots.\]
Well, we see the powers of $2$ on display, all strung out on a line!

If we change the fraction to $1/99997$ we get something just as nice:
\[\frac{1}{99997} = 0.00001 \ 00003 \ 00009 \ 00027 \ 00081 \ 00243 \ 00729 \ \ldots.\]
Now we see the powers of $3$. Isn't that a pretty sight?

Once we get the general idea, it isn't at all hard to generalize it. For example, here is $1/9995$:
\[\frac{1}{99995} = 0.00001 \ 00005 \ 00025 \ 00125 \ 00625 \ 03125 \ 15625 \ldots.\]
But I hear you jumping up and saying something urgently to me:

"Hold on, now! These are the decimal expansions of rational numbers, so they ought to yield recurring decimals, right? But where are the recurring sequences in the above decimals? I don't see any!" 

Well, I'll leave it to you to figure that one out!

One can even get the Fibonacci numbers strung out on a line:
\[\frac{1000}{998999} = 0.001 \ 001 \ 002 \ 003 \ 005 \ 008 \ 013 \ 021 \ 034 \ 055 \ 089 \  \ldots.\]
Try to figure out which fraction will yield the Lucas numbers.

(The Lucas sequence is less well known than the Fibonacci sequence, but is defined in a very similar way. Its starting numbers are $2$ and $1$, in that order, and following these, each subsequent number is the sum of the two that precede it
\[2, \,\ 1, \,\ 3, \,\ 4, \,\ 7, \,\ 11, \,\ 18, \,\ 29, \,\ 47, \,\ 76, \,\ \ldots.\]
These numbers have extremely close interconnections with the Fibonacci numbers.)

\[\clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit\]

Primes in arithmetic progression

How many primes can you find in arithmetic progression (prime AP for short)? Recall that in an AP, the differences between successive members are the same, so the AP has the form $a$, $a+d$, $a+2d$, $a+3d$, $a+4d$, .... Here $a$ is called the "first term" while $d$ is the "common difference". For example, $1$, $4$, $7$, $10$ is a four term AP with common difference $d = 3$.

Here are some examples of prime APs:

• {$3$, $5$, $7$}, with length $3$ and common difference $2$;
• {$5$, $11$, $17$, $23$, $29$}, with length $5$ and common difference $6$;
• {$7$, $37$, $67$, $97$, $127$, $157$}, with length $6$ and common difference $30$.
• {$7$, $157$, $307$, $457$, $607$, $757$, $907$}, with length $7$ and common difference $150$.

It is easy to check that the prime AP {$3$, $5$, $7$}, with length $3$ and common difference $2$, is the only such prime AP, and a prime AP with common difference $2$ cannot exceed a length of $3$. To see why, we only need to examine the remainders left by these numbers on division by $3$.

It is also easy to see the following:

• If a prime AP has length greater than $2$, then its common difference must be an even number, and it must feature only odd primes (because $2$ is the only even prime).
• If a prime AP has length greater than $3$, then its common difference must be a multiple of $3$, and therefore a multiple of $6$ as well (since it also must be an even number). To see why, we only need to examine the remainders left by these numbers on division by $3$.
• If a prime AP has length greater than $5$, then its common difference must be a multiple of $5$, and therefore a multiple of $30$ as well (since it also must be a multiple of $6$). To see why, we only need to examine the remainders left by these numbers on division by $5$.
• If a prime AP has length greater than $7$, then its common difference must be a multiple of $7$, and therefore a multiple of $210$ as well (since it also must be a multiple of $30$). To see why, we only need to examine the remainders left by these numbers on division by $7$.

This chain of results may clearly be extended indefinitely.

By these means we can list innumerable necessary conditions for the existence of long prime APs.

But finding such APs is quite a different matter. As of now, it is largely based on trial-and-error, but guided by clever heuristics. In other words: try, try, and try again! To get a sense of the difficulties involved in this, try finding a prime AP of length $10$ or more - on your own!

Recently (earlier this month, in fact), researchers found a prime AP of length $26$. Here is a link to the site where the discovery was announced: AP26 discovery. The first prime of this AP is
\[43142746595714191,\]
and its common difference is this number:
\[23681770 \times (2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23).\]
In other words, this AP consists of the numbers
\[43142746595714191 + 5283234035979900 \times n\]
for $n = 0$, $1$, $2$, ..., $25$.

Note that the number
\[2 \times 3 \times 5 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 = 223092870,\]
which is a factor of the common difference, is simply the product of the first $9$ primes. (You will find numbers of this kind repeatedly featuring in such searches.) You can check that the last member of this prime AP is the number
\[175223597495211691.\]
To verify that all these $26$ numbers are really prime would require a good computer algebra system, like Derive or Mathematica or the free open source package, Maxima. Here is what Maxima tells you when you ask it whether the $27$-th term of the AP is prime or not - that is, the number
\[43142746595714191 + 5283234035979900 \times 26;\]
Maxima returns the factorization
\[29 \times 31 \times 41 \times 59 \times 85433250011.\]
So this particular AP cannot be extended beyond $26$ terms.

As of now, this is the "champion" - the holder of the "world record"! No prime AP of length greater than $26$ is currently known.

But just a few years back, Terence Tao and Ben Green proved that there exist prime APs of any desired length! (We had featured Terence Tao in a previous posting.) In other words, there must exist a prime AP of length exceeding one billion; only, we may never ever find it, for it may well involve stupendously large numbers ...

Here are two other sites which will tell you a lot about this topic: Prime AP records and Wikipedia-primes in AP.
\[\clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit \quad \clubsuit\]

Views from Terence Tao

Here is an interesting interview with the remarkably prolific Terence Tao, who is one of the youngest Field Medalists ever (he got the Medal in 2006, during the last ICM); it is certainly worth reading:

• Terence Tao - interview

And here is Tao writing on his own blog, on a topic much discussed these days: the potentialities of the internet in teaching; this is actually a talk given at the American Academy of Arts and Sciences, at their induction ceremony:

• Speech given at American Academy of Arts and Sciences

What is interesting is that he posted a rough draft of the talk on his blog, prior to the actual talk, and invited comments and feedback, and then incorporated some of what he received into his talk.

Today, in the light of radical changes currently happening in the Indian education system, such as the passing of the Right to Education Act, there is considerable talk going on about teacher education. (There is a gigantic shortage in the supply of trained teachers today.) A key question here is whether Web based delivery systems have any role to play in tackling this vast problem. I think Prof Tao's comments have value in this context. A lot of thought needs to be given to this question, and I invite readers to share their views and post their comments. Links to pieces written about this would be very welcome.

Big numbers!

Here is a delightful piece about big numbers, written by Professor Marcus du Sautoy, who is a mathematician and a remarkably gifted expositor, and Oxford University's Simonyi Professor of the Public Understanding of Science (a post held till recently by Professor Richard Dawkins):

• Big numbers

The piece is from the Guardian (published in the UK). If you want to read more about the author, here is another piece from the Guardian:

• Marcus du Sautoy

It not tells you about the author but also describes some of the challenges he faces in today's cultural context.

Teachers will be particularly interested in this essay by du Sautoy, on how to spark off interest in mathematics in young children; it too is from the Guardian:

• Sparking interest in maths

It is, in fact, a must-read for all mathematics teachers.

Almost Pythagorean Triples (APTs) - 1

Practically everyone is familiar with Pythagorean triples - triples $(a,b,c)$ of positive integers for which
\[a^2 + b^2 = c^2.\]
If $a, b, c$ are coprime we call the triple a Primitive Pythagorean Triple, or PPT for short. Well known examples are the triples $(3, 4, 5)$ and $(5, 12, 13)$. There is a huge literature about PPTs: how we can generate them in a systematic way, what kinds of number theoretic properties they possess, and so on. Here is a typically striking result about PPTs:

• If $(a, b, c)$ is a PPT, then $a b c$ is divisible by $60$.

Here are some links to material about PPTs available on my own site, MathCelebration:

• PPT1-screen
• PPT1-print
• PPT2-screen
• PPT2-print

Let us now tweak the problem a bit, and ask about triples of positive integers that almost satisfy the condition required of a Pythagorean triple, missing it by the smallest possible margin (namely, by $1$). Thus, we want triples $(a, b, c)$ of positive integers such that
\[a^2 + b^2 = c^2 \pm 1.\]
If the positive sign holds we shall call $(a, b, c)$ an APT - short for "almost Pythagorean triple". And if the negative sign holds we shall call $(a, b, c)$ a NPT - short for "near Pythagorean triple".

But finding APTs and NPTs turns out to be trickier than finding PPTs. The difficulty is caused by the fact that the defining equations are not homogeneous. In the case of PPTs the defining equation is homogeneous of degree $2$; if we divide the equation $a^2 + b^2 = c^2$ by $c^2$, and write $x = a/c$, $y = b/c$, we get the equation $x^2 + y^2 = 1$, whose form immediately brings to mind the equation of a unit circle in the Cartesian plane. This observation if pursued yields a parametric solution to the Pythagorean equation. It is just this approach that has been followed in the articles listed above.

But this approach fails in the case of APTs and NPTs; the non-homogeneity of the equation $x^2 + y^2 = z^2 \pm 1$ negates all such attempts. So for the moment let me not say how we can go about finding them, but simply exhibit a few such triples - found simply by "brute force search" using a computer.

Here are some APTs. For simplicity I have taken $a \le b \le 30$.

a    b    c

_________

4     7     8 

5     5     7 

6    17   18 

7    11   13 

8     9   12 

9   19   21 

10   15   18 

11   13   17 

11   29   31 

13   19   23 

14   17   22 

15   26   30 

16   23   28 

17   21   27 

19   27   33 

20   25   32 

23   29   37 

29   29   41 

(Comment, in passing. The table looks terrible from a typesetting point of view, doesn't it? But I have yet to figure out how to typeset LaTeX-style tables in blogger.)

Try generating more such triples, and look for patterns hidden in them. I think there are more patterns there than one may expect.

The same thing is true for NPTs - i.e., triples $(a, b, c)$ of positive integers for which $a^2 + b^2 = c^2 - 1$. Here are a few such triples, generated the same way (with $a \le b \le 30$):

a   b   c

________

2    2    3  

4    8    9  

6   18  19  

12   12   17  

18   30   35  

If you experiment on this problem on your own, you will hit upon several oddities which are difficult to explain. For example, there seem to be many more APTs than NPTs. Within the region $a \le b \le 30$, the above two tables list all APTs and all NPTs. Observe that the APTs outnumber the NPTs by a fairly big margin. This discrepancy persists as the upper bound is raised. It is not at all clear why this is so.

We'll continue on this theme in the next post.

Jaime Escalante

Jaime Escalante, a mathematics teacher who taught in Los Angeles, California, USA, under very difficult conditions and achieved something truly remarkable, has just passed away (on 30 March 2010). Please see these two links:

• Jaime Escalante - Wikipedia, the free encyclopedia.
• LA Times - obituary

There was even a film made about him, in 1988 - Stand And Deliver.

It would be wonderful if we could document similar stories from India. I am certain that there are many, many unsung heroes out there, in the small town and villages, unknown to all except those who studied under them and learned not only the beauty of our subject but also about the vast canvas of life.

A curious property of the primes

The primes numbers hold many, many secrets --- but unfortunately for us, they guard these secrets a little too efficiently. Most of the time, when we observe a pattern in the sequence of primes, we are unable to find a satisfactory explanation or proof.

But once in a while we meet a property which is amenable to a simple proof. Here is one such property.

Let $p_n$ denote the $n$-th prime number. The sequence of primes is
\[\left\langle p_n \right \rangle_{n=1}^{\infty} \ = \ \left \langle 2, \, 3, \, 5, \, 7, \, 11, \, 13, \, 17, \, 19, \, 23, \, 29, \, 31, \, \ldots\right\rangle.\]
Let us form the partial sums of this sequence, i.e., the sums

\[2, \, 2+3, \, 2+3+5, \, 2+3+5+7, \, 2+3+5+7+11, \, \ldots.\]

We get the following sequence of numbers, which we denote by $\left \langle a_n \right \rangle_{n=1}^{\infty}$, so that $a_n = p_1 +p_2 + \cdots + p_n = \sum_{i=1}^n p_i$:

\[\left\langle a_n \right\rangle_{n=1}^{\infty} \ = \ \left\langle 2, \, 5, \, 10, \, 17, \, 28, \, 41, \, 58, \, 77, \, 100, \, 129, \, 160, \, \ldots\right\rangle.\]
Here is an interesting property of the above sequence of partial sums:

• Between every two numbers in the sequence there exists a perfect square number. 

In other words, for each positive integer $n$, there exists an integer $k$ such that
\[a_n < k^2 < a_{n+1}.\]
Note that we are not asserting that there is just one such number: there may well be more than one square number between $a_n$ and $a_{n+1}$. Indeed, a counterexample to such a claim (were anyone to make it) is easily spotted: we find that $a_{11} = 160$ and $a_{12} = 197$, and between the two numbers $160$ and $197$ there lie two square numbers, $169$ and $196$.

Proof, anyone?

Proof of the Arbelos Theorem - 2

I have written a computational proof of the arbelos theorem and uploaded it to the "Articles" page of my website, MathCelebration. Here are the exact URLs of the two files (though clicking on them will open the documents in Google Docs):

• Arbelos proof - screen version
• Arbelos proof - print version

If you glance through the proof, you will see that it is highly computational, but it uses nothing more advanced or complicated than the theorem of Pythagoras.

The link I gave earlier takes you to the original "pure geometry" proof given by Archimedes, but it shows the congruence only of $\omega_4$ and $\omega_5$. The above proof establishes that $\omega_7$ too has the same radius.

I close with a rather challenging question:

Given the configuration with circles $\omega_1$, $\omega_2$ and $\omega_3$, how will you geometrically construct circles $\omega_4$, $\omega_5$, $\omega_6$ and $\omega_7$? 

(Here, "geometrically" means that we work only with Euclidean instruments.)

Graphics in LaTeX

In today's post I will say something about the graphics capabilities that one can access from within LaTeX.

Two distinct approaches are possible:

• the diagram to be inserted may be prepared using some external application (this includes the case when the diagram is a photograph, so it is very useful when we have to insert photographs into the document); 
• or the commands needed to generate the diagram may be included in the LaTeX file itself, in which case the diagram is generated in real time ("on the fly"), as the document is being processed.

LaTeX allows both these approaches. The former option is used when the graphic to be inserted is available as a graphics file (a jpeg or gif or png or eps file). The command used is something like:

\includegraphics[various options, including desired size of diagram]{name of file with path and extension}

The options allow resizing in various ways, cropping, etc. Details on how this is done can be found in any of the help documents listed in the earlier post. (Note. In LaTeX, options are generally placed within square brackets.)

Here I propose to say something about the other situation - when the diagram is drawn in real time.

LaTeX itself has an inbuilt picture editor whose usage is as shown below:

\begin{picture}
... various commands to draw lines, polygons, dots, circles, etc ...
\end{picture}

This is adequate for simple pictures; but it is really very rudimentary in its scope. (I myself never use it, having gotten hooked onto PSTricks early in my LaTeX days.)

That brings me to PSTricks, which in contrast is highly sophisticated and customizable. Precisely because it has been found to be so good by so many users, the community of people working on PSTricks is vast, and new modules (packages) are constantly being released; the basic package itself is constantly undergoing improvements by its authors (Timothy Van Zandt, Denis Girou, Sebastian Rahtz and Herbert Voss).

A typical environment within which a diagram is drawn using PSTricks looks like this:

\begin{pspicture}

... various commands to draw lines, polygons, dots, circles, etc ...

\end{pspicture}

Here are some links for finding out more about PSTricks:

• Wikipedia-PSTricks
• TUG-PSTricks
• TUG-PSTricks-documents
• TUG-PSTricks-examples
• PSTricks-quick ref list
• PSTricks-user guide
• PSTricks-full document
• PSTricks-presentation

This list should suffice to get you started. Please examine these links - you will find PSTricks to be a really marvelous device for drawing complex diagrams. 

The best thing is that the package is still growing in usability and versatility. There is a very useful mailing list to which a serious user should certainly subscribe; here is its URL:

• PSTricks-mailing list

If you post queries there, you will find answers coming in from all parts of the world.

Do try it out!

Proof of the Arbelos Theorem

Recently I posted an account of the beautiful Arbelos Theorem found over two thousand years back by the great Archimedes of Syracuse. (See Archimedes's Arbelos Theorem.) Here I give a link to the original proof given by Archimedes:

• Proof by Archimedes of Arbelos Theorem

It is a "pure geometry" proof - it involves almost no computation, and it skillfully uses results from circle geometry and the geometry of homothetic transformations. (Comment. The word "homothety" is of relatively recent origin, but the notion of a homothety or enlargement about a point is an old one.)

However it establishes only that circles $\omega_4$ and $\omega_5$ have equal radius. I do not know any such proof that circle $\omega_7$ too has the same radius.My proof is computational, and I will post it shortly.

"Pattern of Signs" - two uploads

I have just posted two pdf files concerning the "Pattern of Signs" post I made earlier in March (here is the link: Pattern of signs), in which I prove the claims and explain the patterns noted in that post. Here are the links to the two files (the contents of the two files are the same, but one is suitable for reading on the screen, and the other one is suitable for printing):

• Pattern of signs - screen version
• Pattern of signs - print version

Actually, these links lead to the Google Docs versions of the two articles, but the articles can be downloaded as pdf files by clicking on an appropriate button at the top.

I look forward to observations from readers.

A sample LaTeX document - 2

I have uploaded another LaTeX generated document to Google Docs; here is its link: 

• Arbelos theorem

The document described the Arbelos Theorem of Archimedes - a theorem dating from Greek times, but with a modern component, discovered less than half a century back.


The figure has been drawn using PSTricks, and the source file too is here for readers to look at:

• Arbelos theorem tex file

Do check it out!


(Actually, I tried to upload a Java-enabled "live" figure, which the viewer can manipulate, but Google Docs refused to accept the upload. Or rather, I couldn't figure out a way of doing such an upload. Does anyone know how this can be done?)

A sample LaTeX document

In my last post I made a few remarks about LaTeX and gave several links, some to sites from where the software could be downloaded, and some to LaTeX help sites and tutorials.


Just to show how a typical TeX file looks, and what kind of output it can yield, I am now posting links to some documents which I just uploaded to Google Docs (and therefore residing somewhere in the clouds). The first link leads to the pdf version of an article I recently published in Resonance, the second link leads to the LaTeX file of the article, and the third link leads to the ps (PostScript) version; it can be viewed using GhostView (or Evince, if you have a Linux based system).

• Madhava-pdf
• Madhava-LaTeX
• Madhava-ps

Observe that the TeX file is much lighter than the other two files (just 18 K, as compared with 53K and 228K, respectively). Moreover, the TeX file is a text file (ASCII format) which can be opened by any text editor whatever; so it is platform independent.

The figure drawn on page 3 of the article has been done using PSTricks. If you look within the TeX document you will find the code for the figure in lines 89 through 154. All the commands needed to draw the figure are contained in that code. 

Typesetting of mathematical text using LaTeX

I have often been asked about the LaTeX typesetting software. I will make a few comments here, and provide some links.


LaTeX is Open Source as well as free software. This has the consequence that it benefits from work done all over the world by extremely well intentioned people who have an interest in its development. There are numerous mailing lists, for example:

• TUGIndia - see TUGIndiaMailingList
• Two Google groups, comp.text.tex and LaTeX users group

These lists are very helpful, with subscribers all over the world willing to give of their time to help you out when you are in doubt.


In working with LaTeX you actually work on a text editor. I use TeXnic Center, but there are many other editors available. See FreeTexEditors. The best and easiest to use, in my view, are: 

• TeXnic Center (see Wikipedia-TeXnicCenter and TexnicCenter) 
• Texmaker (see Wikipedia-Texmaker and Texmaker)
• Texworks (see Texworks)

They are all freely downloadable.


Note that LaTeX is not WYSIWYG software. When you work on the text editor, all you will see along with the main text are a lot of markup tags - dollar symbols, backslashes, etc. It is only when the LaTex file is processed ("LaTeXed" as a LaTeX user would say) that you will see the typeset output.


Because of this, and because so many of us have been brought up on WYSIWYG software of various kinds, the initial learning curve for LaTeX is a bit steep. But it is important that we do not give up our experiments with LaTeX at this stage, because the effort will be more than worth it. There is nothing like LaTeX! You will appreciate this when you see the quality of the typeset text.


Assuming you are using Windows I recommend going in for MikTeX. Please go to MikTeX and download the entire package (the current version is MikTeX 2.8). The full instructions will be found on that webpage, for the download, installation. Then download one of the editors listed above, and configure it to MikTeX. Note that MikTeX has an inbuilt updating mechanism.


Here are some LaTeX tutorials freely available on the net (there are many more of this kind, please look on the net):

• Duke tutorial
  
• TUGIndia LaTeX tutorials
  
• LaTeX-wiki
  
• LaTeX tutorial
  

See also:

• LaTeX project
  
• ProTeXt
  

The last link (ProTeXt) is particularly convenient: it allows you to download a single archive that installs all the packages you need - MikTeX, Texnic Center, Ghostscript, Ghostview, etc. I myself did my entire LaTeX related installation using ProTeXt.


Next time I will say something about graphics editors that work with LaTeX.

Tests of divisibility - 3

A small variation in the tests described in the last two posts yields more findings.


Instead of the mapping $10a + b \mapsto a - k b$, what if we have $10a + b \mapsto a + k b$ (that is, with a plus sign replacing the minus sign)? 


Nearly the same steps lead us to see that this yields a test for divisibility by $10k - 1$.


Thus, we consider the following function $g_k$ which acts on the positive integers $x$ as follows: If $x = 10a + b$, where $0 \le b \le 9$, then $g_k (x) = a + k b$. Then we claim that $x$ is divisible by $10k - 1$ if and only if $g_k (x)$ is divisible by $10k - 1$.


To see that this claim is valid, observe that
\[k x - g_k (x) = k(10a + b) - (a + kb) = a(10k - 1).\]
Thus, $k x - g_k (x)$ is a multiple of $10k - 1$. Now:

• If $x$ is divisible by $10k - 1$, then so is $k x$, and so also is $g_k (x)$ from the above relation.
• If $g_k (x)$ is divisible by $10k - 1$, then so also is $k x$, from the above relation. And since $k$ and $10k - 1$ are coprime, this means that $x$ itself is divisible by $10k - 1$.

It follows that $x$ is divisible by $10k - 1$ if and only if $g_k (x)$ is divisible by $10k - 1$. This proves our claim.


Just like earlier we get several tests of divisibility from this fact, all at the same time.

• Putting $k = 2$, we get a test of divisibility by $19$: The integer $10a + b$ is divisible by $19$ if and only if $a + 2b$ is divisible by $19$. So the test is: Add twice the units digit to the rest of the number, and proceed as earlier.
• Putting $k = 3$, we get a test of divisibility by $29$: The integer $10a + b$ is divisible by $29$ if and only if $a + 3b$ is divisible by $29$. So the test is: Add three times the units digit to the rest of the number, and proceed as earlier.
• With $k = 4$ we get a test for divisibility by $13$ (which is a divisor of $39$), namely: The integer $10a + b$ is divisible by $13$ if and only if $a + 4b$ is divisible by $13$. So the test is: Add four times the units digit to the rest of the number, and proceed as earlier.
• Continuing, $k = 5$ yields another test for divisibility by $7$ (because $49 = 7 \times 7$), and $k = 6$ yields a test for divisibility by $59$. 
• Of particular interest is the case $k = 10$, for then $10k - 1 = 99$, which factorizes as $9 \times 11$. So this is simultaneously a test for divisibility by $9$ as well as by $11$. The test is: Add ten times the units digit to the rest of the number, and proceed as earlier. Then the original number is divisible by $9$ if and only if the final number is divisible by $9$, and the original number is divisible by $11$ if and only if the final number is divisible by $11$.

Quite a rich set of findings, all of which emerge from such a simple observation!

Tests of divisibility - 2

I asked last time, "What is the significance of $2$ as a multiplier in the test for divisibility by $7$?'' I also asked, ``Are there tests similar to this for testing divisibility by other divisors - like $9$, $11$, $13$, $17$, $19$, ..."


The answer to the latter question is a loud "Yes!". Indeed, one can find such a test for any divisor which has no factors in common with $10$; that is, any divisor whose units digit is $1$, $3$, $7$ or $9$.


The logic behind this gets revealed when we understand the role played by the multiplier $2$ in the test for divisibility by $7$.


One way to understand the role played by $2$ is to note that $7$ has $21$ as a multiple (note that $21$ has $2$ as its tens digit), and if we apply the transformation $10a + b \mapsto a - 2b$ to $21$, we get $0$ right away. We also get $0$ if we apply it to multiples of $21$ like $42$, $63$, $84$, $105$, $126$, ... Try it out and you'll see this for yourself.


Noting this, we are led to invent a new test of divisibility. Let $k$ be any positive integer. Consider the following function $f_k$ which acts on the positive integers $x$ thus: If $x = 10a + b$, where $0 \le b \le 9$, then $f_k (x) = a - k b$. Then:

• $x$ is divisible by $10k + 1$ if and only if $f_k (x)$ is divisible by $10k + 1$.

To see that this claim is valid, observe that
\[kx + f_k (x) = k(10a + b) + (a - kb) = a(10k + 1).\]
Now:

• If $x$ is divisible by $10k + 1$, then so is $kx$, and so also is $f_k (x)$ from the above relation.
• If $f_k (x)$ is divisible by $10k + 1$, then so also is $kx$, from the above relation. And since $k$ and $10k + 1$ are coprime (they must be, because $10k + 1$ leaves remainder $1$ when divided by $k$), this means that $x$ itself is divisible by $10k + 1$.

It follows that $x$ is divisible by $10k + 1$ if and only if $f_k (x)$ is divisible by $10k + 1$.


From this observation we quickly get many different tests of divisibility, all at the same time (in fact, infinitely many of them):

• Putting $k = 3$, we get a test of divisibility by $31$: The integer $10a + b$ is divisible by $31$ if and only if $a - 3b$ is divisible by $31$. So the test is: Subtract three times the units digit from the rest of the number, and proceed as earlier.
• Putting $k = 4$, we get a test for divisibility by $41$: The integer $10a + b$ is divisible by $41$ if and only if $a - 4b$ is divisible by $41$. So the test is: Subtract four times the units digit from the rest of the number, and proceed as earlier.
• Similarly, $k = 6$ gives a test for divisibility by $61$, $k = 7$ gives a test of divisibility by $71$, and so on.

What about $k = 2$? This gives us a test for divisibility by $21$. But $21 = 3 \times 7$ is a composite number. So the same test serves as a test for divisibility by both $3$ and $7$. Hence:

• The number $10a + b$ is divisible by $3$ if and only if $a - 2b$ is divisible by $3$.
• The number $10a + b$ is divisible by $7$ if and only if $a - 2b$ is divisible by $7$.

Similarly, if $k = 5$ we get another composite number, $51 = 3 \times 17$. Hence:

• The number $10a + b$ is divisible by $17$ if and only if $a - 5b$ is divisible by $17$.

(Note that from $k = 5$ we get another test for divisibility by $3$; however a bit of thought will show that it is equivalent to the test obtained by taking $k = 2$. So we need not list it here.)


If $k = 9$ we get yet another composite number, $91 = 7 \times 13$. So from this we get a test for divisibility by $13$:

• The number $10a + b$ is divisible by $13$ if and only if $a - 9b$ is divisible by $13$.

Two values of $k$ of particular interest are $k = 1$, and $k = 8$. Let us see why.


If $k = 1$ then $10k + 1 = 11$. The test now is:

• The number $10a + b$ is divisible by $11$ if and only if $a - b$ is divisible by $11$.

Now the mapping $10a + b \mapsto a - b$ amounts to this: Subtract the units digit from the rest of the number.


I wonder if you see that this test is just a disguised form of the usual test for divisibility by $11$.


If $k = 8$ then $10k + 1 = 81 = 9 \times 9$; so this should give us a test for divisibility by $9$. The test is:

• The number $10a + b$ is divisible by $9$ if and only if $a - 8b$ is divisible by $9$.

Now, it is obvious that $a - 8b$ is divisible by $9$ if and only if $a + b$ is divisible by $9$ (because $(a + b) - (a - 8b) = 9b$, which is a mutiple of $9$). Hence:

• The number $10a + b$ is divisible by $9$ if and only if $a + b$ is divisible by $9$.

Now the mapping $10a + b \mapsto a + b$ amounts to this: Add the units digit to the rest of the number.


I wonder if you see that this test too is just a disguised form of the usual test for divisibility by $9$.


In this manner we can get a test for divisibility by any number with units digit $1$, and by any number which divides a number with units digit $1$.


A very similar line of reasoning gives tests for divisibility by numbers with units digit $9$; i.e., divisors like $19$, $29$, .... But we'll leave this to the next post.

Tests of divisibility - 1

Everyone knows the tests for divisibility by $2$ and $5$; they are obvious. The tests for divisibility by $4$, $8$, $3$, $6$, $9$ and $11$ are less obvious but still well known; each one is based in some way on the digits of the given number. (The logic behind these tests may be less clear - particularly for divisibility by $11$. But we expect that most students have at least an inkling of the underlying logic. I hope so anyway!) 

What about divisibility by $7$? Some of you may know of this test: 

Given an integer $N$, let $b$ be its units digit, and let $a$ be the rest of the number, obtained by deleting the units digit; this means that $N = 10a + b$. Now replace $N$ by $a - 2b$. In other word, subtract twice the units digit from the rest of the number. Now repeat the same step for the new number, and keep doing this till you get a small number $M$ for which you can check divisibility by $7$ mentally. If this check works out (i.e., $M$ is divisible by $7$), then the original number $N$ is divisible by $7$; else it is not. 

Thus this is an ``if and only if'' test. Here are two examples:

• If $N = 123456$, then we do $12345 - 12 = 12333$. Next we do: $1233 - 6 = 1227$. Next: $122 - 14 = 108$, and $10-16 = -6$. The last number ($-6$) is clearly not a multiple of $7$; hence, neither is the original number, $123456$.
• Or try $N = 1504055$. We get: $150405 - 10 = 150395$. Next: $15039 - 10 = 15029$. Next: $1502 - 18 = 1484$. Next: $148 - 8 = 140$. We can stop now, since $140$ is quite visibly a multiple of $7$. We conclude that $1504055$ too is a multiple of $7$.

Why does this work? Here is a proof; it uses two simple facts: 

• The sum and difference of two multiples of $7$ are also multiples of $7$.
• If $y$ is an integer such that $3y$ is a multiple of $7$, then $y$ itself is a multiple of $7$. (Do you see why? The point is that $3$ and $7$ are relatively prime to each other.)

Let us now show that for any two integers $a$ and $b$, the quantities $x = 10a+b$ and $y = a-2b$ are either both divisible by $7$ or both indivisible by $7$. For we have:

\[x - 3y = (10a + b) - 3(a - 2b) = 7a + 7b = 7(a + b),\]

which tells us that $x-3y$ is a multiple of $7$. Now:

• If $x$ is a multiple of $7$, then $3y$ is a multiple of $7$, and hence so is $y$.
• And if $y$ is a multiple of $7$, then so is $3y$, and hence so is $x$.

This reasoning carries right through the entire working of the algorithm. 


Conclusion. The initial number is a multiple of $7$ if and only if the final number is a multiple of $7$.

So this is why the test works, but the proof raises several more questions:

1. What is the significance of $2$ as a multiplier in the above test (i.e., "twice the units digit")? How could we have found this multiplier?
2. Are there tests similar to this for testing divisibility by $9$? $11$? $13$? $17$? $19$?

We'll try to answer these in the next entry! Cheers till then.

The problem of the four 2's

I am basing today's entry on an entry I saw in another blog: Kolipakkam's Is there a PAM Dirac amongst us?  This in turn was brought to my attention by my colleague and good friend B Sankararaman of The Valley School.


A type of puzzle frequently seen is:

• Using exactly four $2$'s (no less and no more), make the number $5$ (or $6$ or $10$ or $50$ any other given number).

Or:

• Using exactly four $4$'s (no less and no more), make the number $5$ (or $6$ or $10$ or $50$ any other given number).

Or:

• Using the digits $1$, $9$, $4$, $7$ exactly once each (no less and no more), make the number $5$ (or $6$ or $10$ or $50$ any other given number).

Obviously one can make an unlimited number of such puzzles. A popular pastime is to do this while on the road, using the license plate number of the car or bus just ahead of you. Or one can use it to set an exercise for one's students, using the digits of a particular year. This can be a lot of fun at, say, the 7th or 8th standard level.


Here I will make some remarks on the first problem (where we try to make the successive positive integers using just four $2$'s). A lot of fun can be had generating the answers:

• $$1 = \frac{2}{2} \times \frac{2}{2}$$
• $$2 = \frac{2}{2} + \frac{2}{2}$$
• $$3 = 2^2 - \frac{2}{2}$$
• $$4 = 2^2 \times \frac{2}{2}$$
• $$5 = 2^2 + \frac{2}{2}$$
• $$6 = 2 + 2 + \sqrt{2 \times 2}$$
• $$8 = 2^2 + 2^2$$
• $$9 = \left(2 + \frac{2}{2}\right)^2$$

One can go on like this, generating the positive integers one after another (but note that we skipped the entry for $7$). At some point one is bound to ask, 

• Can this go on for ever? Can we express every integer using precisely four $2$'s? Or are there some integers for which the representation cannot be found?

Unfortunately this problem has been solved once for all! - there is a simple algorithm which yields the answer for every positive positive integer $n$. Here is how it works. Let

\[x = 2^{1/2^n}.\]

Note that $x$ depends on $n$, and can be written using just one $2$ and $n$ square root signs:

\[x = \sqrt{\sqrt{\sqrt{ \cdots \sqrt{2}}}},\]

with $n$ square root signs nested inside each other. By taking logarithms to base $2$ we get:

\[\log_2 x = \frac{1}{2^n}.\]

Inverting this relation we get $2^n = 1/\log_2 x$, and hence:

\[2^n = \log_{x} 2.\]

Now take logarithms to base $2$ yet again; we get:

\[n = \log_2 \left( \log_{x} 2 \right).\]

Now putting in the expression we had for $x$ (recall that it used just one $2$ and $n$ square root signs), we get an expression for $n$ using just three $2$'s!

For example, for $n = 5$ we have:

\[5 = \log_2 \left( \log_{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}} 2 \right).\]
And for $n = 7$:

\[7 = \log_2 \left( \log_{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}}}} 2 \right).\] 
Wow!

Note that this solution uses just three $2$'s. But we can very easily convert this to a solution using four $2$'s by (wastefully) writing one of the $2$'s using two $2$'s, say as $\sqrt{2 \times 2}$.


I am told that this solution was given by the great physicist Paul Dirac (who in spirit was a mathematician rather than a physicist; or maybe I should say that he liked to think of the world in purely mathematical terms rather than in physical terms). The story is told more fully in the blog I referred to above (Kolipakkam's Is there a PAM Dirac amongst us?). Do have a look at it.


Why do I call this "unfortunate"? Well, you don't have to take that seriously! On the one hand, Dirac's solution is neat and beautiful. At the same time it is always a matter of regret when a problem gets solved so completely that there is nothing for anyone to add any more ...! Would you agree?

A pattern of signs

Take any four consecutive integers $a, b, c, d$. Then it is obvious that $a+d = b+c$. Let us write this relation as 
\[+a-b-c+d = 0.\]
Note the sequence of signs on the left side: $+,-,-,+$. We shall write the above statement in the following form: 

The sign sequence $+--+$ if applied to any $4$ consecutive integers yields a sum of $0$.

For example, if the 4 consecutive integers are $12$, $13$, $14$, $15$ we have: 
\[+ 12 - 13 - 14 + 15 = 0.\]
Now reverse all the signs in this string; we get the string $-++-$. If we concatenate the two strings together we get the string $+--+-++-$ (which is now of length $8$). Now here is a surprising fact:

The sign sequence $+--+-++-$ if applied any $8$ consecutive squares yields a sum of $0$.

For example, take the $8$ consecutive squares $9$, $16$, $25$, $36$, $49$, $64$, $81$, $100$. We have now:
\[+ 9 - 16 - 25 + 36 - 49 + 64 + 81 - 100 = 0.\]
Please verify this statement. Then try it out on other collections of $8$ consecutive squares.


Now take the string $+--+-++-$ and reverse all its signs; we get the string $-++-+--+$. If we concatenate these two strings together, we get the string $+--+-++--++-+--+$ (which is now of length $16$). Here is the next surprising fact:

The sign sequence $+--+-++--++-+--+$ if applied any $16$ consecutive cubes yields a sum of $0$.

Here is an example: take the $16$ consecutive cubes $27$, $64$, $125$, $216$, $343$, $512$, $729$, $1000$, $1331$, $1728$, $2197$, $2744$, $3375$, $4096$, $4913$, $5832$. Applying the sign sequence just given to these numbers, you will find you get a sum of $0$. I have not written the full "sum" here as it does not fit into this space properly: $+ 27 - 64 - 125 + 216 - 343 + 512 + 729 - 1000 - 1331 + 1728 + 2197 -2744 + 3375 - 4096 - 4913 + 5832 = 0$. But please check it out.

Then try this out for other collections of $16$ consecutive cubes.


I suppose you will now be able to guess what comes next in this sequence of statements .... 


But why do we have such a pattern?

Update on the 120 degree triangle

In the piece that I uploaded to my website (mathcelebration) a few days back, about a property that triangles with a $120^{\circ}$ angle possess, I made the following remark: 


"The pure geometry proof given above is very finely dependent on the hypotheses, and is difficult to generalize (this is a common feature of many such geometric proofs)."


I have just found that this somewhat off the cuff remark was not justified; in fact I have found a pure geometry proof of the converse proposition of that article.


That is, I have found a "pure geometry" way of showing this proposition: if $\angle A$ is not equal to $120^{\circ}$ then $\angle QPR$ is not a right angle. And to my surprise it is a very easy proof!


Here are the links:

• Triangle120degree2Screen
  
• Triangle120degree2PDF

Do have a look at them!