# Chaitanya's Random Pages

## August 31, 2012

### A Pythagorean triangle from the incentre of another Pythagorean triangle

Filed under: mathematics — ckrao @ 1:00 pm
Tags: ,

A Pythagorean triple is a set of three positive integers $a, b, c$ satisfying $a^2 + b^2 = c^2$. Two examples are $(3,4,5)$ and $(5,12,13)$. These form the side lengths of a right-angled triangle, called a Pythagorean triangle.

Some time ago I found that a (3,4,5) triangle can be found inside a (7,24,25) right-angled triangle as follows.

In this post we will generalise the above figure. That is, in a right-angled triangle with integer side lengths consider a smaller right-angled triangle formed by one of its vertices, the incentre, and the incircle’s point of tangency with a side as shown above. For what side lengths of the orignal triangle does this inner triangle also have integer side lengths? Below is a second example.

To find the general solution we use the fact that primitive Pythagorean triples (those without a common factor) have the following form:

$\displaystyle (m^2-n^2, 2mn, m^2 + n^2), \quad(1)$

where $m$ and $n$ are positive integers with no common factor, $m>n$ and one of them is even. Any other (non-primitive) triple is formed by multiplying each of the three elements by the same positive integer.

Here is an easy way to derive (1) that I found here . We have $b^2 = c^2 - a^2 = (c-a)(c+a)$ from which $b/(c-a) = (c+a)/b = m/n$ where $m$ and $n$ are coprime integers. Hence $n/m = (c-a)/b$ and we have the two equations

$\displaystyle \frac{c}{b} + \frac{a}{b}= \frac{m}{n}$

$\displaystyle \frac{c}{b} - \frac{a}{b}= \frac{n}{m}.$

Adding and subtracting the two equations gives $c/b = (m/n + n/m)/2 = (m^2 + n^2)/2mn$ and $a/b = (m/n - n/m)/2 = (m^2 - n^2)/2mn$. We can then equate numerator and denominator leading to the desired result provided the right sides are reduced. This will be the case if $m$ and $n$ have no common factor and one of them is even (this ensures the numerator $m^2 \pm n^2$ is odd).

Since a line from a vertex of a triangle to its incentre is an angle bisector, we have angles of $\theta$ and $2\theta$ in our inner and outer triangles respectively. Since the inner triangle has side lengths forming a Pythagorean triple we have

$\displaystyle \tan \theta = \frac{m^2 - n^2}{2mn}$ or $\displaystyle \tan \theta = \frac{2mn}{m^2 - n^2}$.

Only one of these is less than 1 since they are reciprocals of each other.

Using the identity $\tan 2\theta = 2 \tan \theta /(1 - \tan^2 \theta)$,

either $\displaystyle \tan 2\theta = \frac{(m^2 - n^2)/(2mn)}{1 - [(m^2 - n^2)/(2mn)]^2} = \frac{4mn(m^2-n^2)}{(2mn)^2 - (m^2 - n^2)^2}$

or $\displaystyle \tan 2\theta = \frac{(2mn)/(m^2 - n^2)}{1 - [(2mn)/(m^2 - n^2)]^2} = \frac{4mn(m^2-n^2)}{(m^2 - n^2)^2 - (2mn)^2}$.

These are equivalent up to sign and we choose the case for which $\tan 2\theta > 0$ (i.e. $2\theta < \pi/2$) depending on whether $m^2 - n^2$ is smaller or larger than $2mn$.

Next we show that the numerator $4mn(m^2-n^2) = 4mn(m+n)(m-n)$ is coprime with the denominator $(m^2 - n^2)^2 - (2mn)^2 = (m^2 - n^2 + 2mn)(m^2 - n^2 - 2mn)$ (which is odd when $m$ and $n$ have different parity).  Suppose an odd prime $p$ divides both $(m+n)$ in the numerator and $(m^2 - n^2)^2 - (2mn)^2 = (m+n)^2(m-n)^2 - (2mn)^2$ in the denominator. Then $p$ divides $(2mn)^2$ and so $p$ divides $m$ or $n$. But $p$ divides $(m+n)$ which implies $p$ is a factor of both $m$ and $n$: contradiction. A similar contradiction can be used to show any odd prime factor of $(m-n)$ cannot divide the denominator.

Next if an odd prime $p$ divides both $m$ in the numerator and $(m^2 - n^2)^2 \pm (2mn)^2$ in the denominator, then $m$ divides $n^4$ implying $p$ divides both $m$ and $n$: contradiction. A similar contradiction can be used to show any odd prime factor of $(m-n)$ cannot divide the denominator. We conclude that the numerator and denominator have no common factors.

From these values of $\tan 2\theta$, the numerator and denominator (possibly times a common multiple) correspond directly to side lengths for the outer triangle and we find the hypotenuse to be $\left([4mn(m^2-n^2)]^2 + [(m^2 - n^2)^2 - (2mn)^2]^2\right)^{1/2} = (m^2 +n^2)^2$. Hence the outer triangle’s side lengths correspond to the following Pythagorean triples:

$\displaystyle \left(k(4mn(m^2-n^2)), k((m^2 - n^2)^2 - (2mn)^2), k(m^2 + n^2)^2\right)$

Hence we find the interesting fact that the hypotenuse of the outer triangle will always be a multiple of a perfect square, as is seen in the two above examples. With this as the side lengths of the outer triangle, the inradius will be $(a + b - c)/2 = k(4mn(m^2-n^2) - 8(mn)^2)/2 = 2kmn(m^2 - n^2 - 2mn)$ (the formula $(a+b-c)/2$ is valid for any right angled triangle with hypotenuse $c$). This inradius corresponds to one of the side lengths of the inner triangle. The three side lengths of the inner triangle will have the form

$\displaystyle \left(k(m^2-n^2)(m^2 - n^2 - 2mn), 2kmn(m^2 - n^2 - 2mn), k(m^2 + n^2)(m^2 - n^2 - 2mn)\right)$

The first diagram above corresponds to $m=2,n=1,k=1$ while the second corresponds to $m=3,n=2,k=1$. The next simplest type would be $m=4, n=1,k=1$ leading to a $(56,105,119)$ inner triangle and an $(161, 240, 289)$ outer triangle.

## August 29, 2012

### London 2012 recap

Filed under: sport — ckrao @ 12:39 pm

Following the delightful London 2012 Olympic Games, here are some memorable performances and links to articles I have collected.

If I had to pick a single event as most memorable, it would be the men’s 800m final. It was the first time all eight runners went under 1:44 and all but one runner set a personal best time (that one runner set a season best). As well as a world record for David Rudisha, silver medallist Nigel Amos won Botswana’s first ever Olympic medal and both he and bronze medallist Timothy Kitum are aged 18 and 17 respectively! The future looks bright in this event! Some excellent analysis of the event is at the Science of Sport blog here.

 Rank Athlete Nationality Time Notes 1 David Rudisha Kenya 1:40.91 WR, OR, NR 2 Nigel Amos Botswana 1:41.71 WJR, NR 3 Timothy Kitum Kenya 1:42.53 PB 4 Duane Solomon United States 1:42.82 PB 5 Nick Symmonds United States 1:43.05 PB 6 Mohammed Aman Ethiopia 1:43.20 NR 7 Abubaker Kaki Sudan 1:43.32 SB 8 Andrew Osagie Great Britain 1:43.87 PB

(NR = national record, WJR = world junior record, table from here)

The 200m splits were: 23.5s, 25.8 (49.28), 25.0 (1:14.3), 26.6 (more details here)

The US  women’s 4x100m relay team smashed the old world record by 0.55 seconds, a record that was set by the German Democratic Republic in 1985! The 100m splits for the four runners were as follows.

• Felix 9.97
• Knight 10.33
• Jeter 9.70

Jeter may have set the fastest relay split ever if this list is anything to go by. Her PB of 10.64 in the 100m is second behind Florence Griffith-Joyner.

Incidentally here are the corresponding splits for the winning Jamaica men’s 4×100 relay (from here):

• Carter 10.28
• Frater 9.07
• Blake 9.09
• Bolt 8.70

Bolt may have equalled the fastest ever relay split by Asafa Powell at the 2008 Beijing Olympics.

Here is a list of track and field athletes who have secured a second gold medal in the same individual event.

• Usain Bolt in the men’s 100m and 200m (the first to repeat in both events). There’s some good information about his splits for the 100m here.
• Ezekiel Kemboi (Kenya) in the 3000m steeplechase (previous gold in 2004)
• Felix Sanchez (Cuba) in the men’s 400m hurdles (previous gold in 2004)
• Tomasz Majewski (Poland) in the men’s shot put
• Shelly-Ann Fraser Pryce (Jamaica) in the women’s 100m
• Meseret Defar (Ethiopia) in the women’s 5000m (previous gold in 2004)
• Tirunesh Dibaba (Ethiopia) in the women’s 10000m
• Barbora Spotakova (Czech Republic) in the women’s javelin throw
• Valerie Adams (New Zealand) in the women’s shot put

In other sports:

• Michael Phelps (USA) became the first male swimmer to win three consecutive individual gold medals and did so twice (100m butterfly, 200m individual medley). For these games he won four gold medals and two silver medals (his first silvers in Olympic competition!) to end a most remarkable career with 18 gold medals, 2 silvers and 2 bronzes in Olympic competition.
• Ben Ainslie (GB) became the first person to win medals in sailing at five Olympic Games, and the second (after Paul Elvstrøm) to win four gold medals. He got his third consecutive gold medal in the Finn class.
• Saori Yoshida and Kaori Icho (both from Japan) each won their third consecutive gold medal in freestyle wrestling (55kg and 63kg weight classes respectively).
• Valentina Vezzali (Italy) won her sixth gold medal in foil fencing (three individual, three team).
• Tony Estanguet (France) won his third consecutive gold medal in the C-1 canoe class (slalom). He is the first Frenchman to achieve three gold medals in the same event.
• Also winning their third gold medal on the trot were the Williams sisters in women’s doubles tennis and Kerri Walsh and Misty May-Treanor in women’s beach volleyball. Serena Williams also won gold in the singles, losing just 17 games in 6 matches and winning against two of the top three seeds with scorelines of 6-1 6-2 (semi-final vs Azarenka) and 6-0 6-1 (final vs Sharapova).
• Cyclist Sir Chris Hoy (GB) picked up two gold medals (Team sprint, Kieren) to become the most successful British Olympian with six gold medals in total.
• Zou Kai (China) picked up two gold medals in men’s gymnastics (floor exercise, team) to hold the record for the most gold medals by any Chinese olympian.
• Wu Minxia (China) became the first female diver to win three consecutive gold medals (synchronised diving). She also won the 3m springboard competition.
• Anastasia Davydova (Russia) won her fifth gold medal in synchronised swimming after winning the team competition for the third time.

There were 32 world records set including 8 in swimming. The youngest two record-setters were Ye Shiwen (China) in the 400m individual medley and Missy Franklin (USA) in the 200m backstroke. The latter ended up with 5 medals, including four gold. A list of other records is at the Guiness World Records site here.

There were some memorable performances by the host country too – Great Britain won a most impressive 29 gold medals (they won only 1 gold in Atlanta in 1996!) including popular wins by Bradley Wiggins (cycling time trial following his Tour de France win), Jessica Ennis (heptathlon), Mo Farrah (5k, 10k) and Andy Murray (in tennis, defeating the top two players along the way).

A couple of wins from unexpected countries – Keshorn Walcott (Trinidad and Tobago) in the men’s javelin and Arthur Zanetti (Brazil) in the men’s rings (gymnastics).

Finally, some infographics related to the Olympics can be found via Pinterest here. The Guardian also ran an interesting series on the data of the games here and the New York Times has a most interesting graphics collection here.

## August 28, 2012

Filed under: mathematics — ckrao @ 2:02 pm

(This post is inspired by this recent post at the Futility Closet blog.)

Here is a pretty cool equation.

$\displaystyle 1830 + 5637 + 6105 = 7218 + 3411 + 2943\quad \quad (1)$

Why is it cool? Permute the digits of each term in the same way and the equation is still true.

e.g. swapping the second and fourth digits gives

$\displaystyle 1038 + 5736 + 6501 = 7812 + 3114 + 2349$

Furthermore, take any subset of the digits of each term and equality still holds.

e.g. taking the first and third digits of each term (1) gives

$\displaystyle 13 + 53 + 60 = 71 + 31 + 24$

Now for the coolest part – all of the above is true when each term of any of the sums is squared!

e.g.

$\displaystyle 1830^2 + 5637^2 + 6105^2 = 7218^2 + 3411^2 + 2943^2$

$\displaystyle 1038^2 + 5736^2 + 6501^2 = 7812^2 + 3114^2 + 2349^2$

$\displaystyle 13^2 + 53^2 + 60^2 = 71^2 + 31^2 + 24^2$

It may not take too long to see why the initial statements are true (i.e. before we started squaring things). The original sum has this form:

$\begin{array}{lcl} & & (1000r_3 + 100 r_2 + 10r_1 + r_0) + (1000 s_3 + 100s_2 + 10s_1 + s_0)\\ & & \quad + (1000 t_3 + 100t_2 + 10t_1 + t_0)\\ & = & (1000 u_3 + 100u_2 + 10 u_1+ u_0) + (1000 v_3 + 100 v_2 + 10 v_1 + v_0)\\ & & \quad + (1000 w_3 + 100 w_2 + 10 w_1 + w_0) \end{array}$

where the variables indicate digits.

As long as we have for each digit $r_i + s_ i + t_i = u_i + v_i + w_i$ for $i = 0, 1,2,3$, the above sum follows by multiplying both sides by powers of 10 and adding.

This is true in our case for indeed we have $1 + 5 + 6 = 7 + 3 + 2, 8 + 6 + 1 = 2 + 4 + 9, 3 + 3 + 0 = 1 + 1 + 4, 0 + 7 + 5 = 8 + 1 + 3$.

As long as these sums appear in the same digits place (units, tens, hundreds or thousands), the digits can appear in any order. This explains why any subset or permutation of the digits of (1) should preserve equality.

Now it’s time to look at why equality should hold when the terms are squared.

Firstly notice that the sums of the squares of respective digits are equal.

i.e.

$\displaystyle 1^2 + 5^2 + 6^2 = 7^2 + 3^2 + 2^2 = 62$

$\displaystyle 8^2 + 6^2 + 1^2 = 2^2 + 4^2 + 9^2 = 101$

$\displaystyle 3^2 + 3^2 + 0^2 = 1^2 + 1^2 + 4^2 = 18$

$\displaystyle 0^2 + 7^2 + 5^2 = 8^2 + 1^2 + 3^2 = 74$

Using the notation $r_i, s_i, t_i, u_i, v_i, w_i$ as before, we have

$\begin{array}{lcl} & & \left(\sum_{i=0}^3 10^i r_i\right)^2 + \left(\sum_{i=0}^3 10^i s_i\right)^2 + \left(\sum_{i=0}^3 10^i t_i\right)^2\\& = & \sum_{i=0}^3 10^{2i} r_i^2 + 2\sum_{i

For this to equal the desired right side, we know $\sum_{i=0}^3 10^{2i} (r_i^2 + s_i^2 + t_i^2) = \sum_{i=0}^3 10^{2i} (u_i^2 + v_i^2 + w_i^2)$, but we also need to show that the cross terms $\sum_{i are equal to $\sum_{i. It turns out that this is indeed the case for our special choice of digits.

For convenience let $e_i$ denote the column vector $(r_i, s_i, t_i)^T$ and let $f_i = (u_i, v_i, w_i)^T$ (the $T$ represents transpose). Then we need to verify the vector dot products $e_i^T e_j = f_i^T f_j$,

where

$\displaystyle e_1 = (1,5,6)^T, \quad f_1 = (7,3,2)^T$

$\displaystyle e_2 = (8,6,1)^T, \quad f_2 = (2,4,9)^T$

$\displaystyle e_3 = (3,3,0)^T, \quad f_3 = (1,1,4)^T$

$\displaystyle e_4 = (0,7,5)^T, \quad f_4 = (8,1,3)^T.$

(For example, $e_1^T e_2 = 8 + 30 + 6 = 44, f_1^T f_2 = 14 + 12 + 18 = 44$.)

Note that each coordinate of $e_i + f_i$ is equal to (2/3) of the sum of the coordinates of $e_i$. In matrix notation this fact can be stated by the relation

$\displaystyle e_i + f_i = \frac{2}{3} S e_i$

where

$\displaystyle S = \left[ \begin{array}{ccc}1 & 1 & 1\\1 & 1 & 1\\1 & 1 & 1\end{array} \right].$

This can be written as $f_i = (\frac{2}{3}S - I)e_i$ where $I$ represents the 3×3 identity matrix.

We may then verify that $S^TS = S^2 = 3S$ and so for $i = 0, 1, 2, 3$,

\begin{aligned} f_i^T f_j &= e_i^T(\frac{2}{3}S-I)^T (\frac{2}{3}S-I) e_j\\ &= e_i^T (\frac{4}{9}S^T S - \frac{4}{3}S + I) e_j \\&= e_i^T (\frac{4}{3}S - \frac{4}{3}S + I) e_j\\ &= e_i^T I e_j\\ &= e_i^T e_j. \end{aligned}

It follows that the squared sums are equal:

$\begin{array}{lcl} & & \left(\sum_{i=0}^3 10^i r_i\right)^2 + \left(\sum_{i=0}^3 10^i s_i\right)^2 + \left(\sum_{i=0}^3 10^i t_i\right)^2\\ &=&\left(\sum_{i=0}^3 10^i u_i\right)^2 + \left(\sum_{i=0}^3 10^i v_i\right)^2 + \left(\sum_{i=0}^3 10^i w_i\right)^2 \end{array}$

and it can be seen that any subset or permutation of the indices {1,2,3,4} could be chosen in each sum for equality to hold.

(Aside: Note the above is also true if instead of $f_i = (\frac{2}{3}S - I)e_i$we had the relation $f_i = (\frac{2}{3}S - P)e_i$, where $P$ is any 3×3 permutation matrix ($P$ operates by swapping any two indices of a vector). The orthogonality of the matrix $(\frac{2}{3}S - P)$ follows from the relations $SP = PS = S$ and $P^2 = I$. The important thing is that the same  permutation $P$ is used for each $i$.)

In summary the following steps were made in coming up with (1).

1. Start with randomly chosen pairs of digits $(1,5), (8,6), (3,3), (0,7)$ and add a third digit to each pair so that their sum is a multiple of 3. In our case we chose $6, 1, 0, 5$ so that the resulting triples have sum $1+5+6 = 12, 8+6+1 = 15, 3+3+0 = 6, 0+7+5 = 12$.
2. Given a triple $e_i^T = (a_i,b_i,3k_i-(a_i+b_i))$ with sum $3k_i$ define the second triple to be $f_i^T = (2k_i-a_i, 2k-b, a+b-k)$. Then it can be verified that $e_i^T e_j=f_i^T f_j$ for any $0 \leq i,j \leq 3$. In our example, we obtain the triples $(8-1,8-5,8-6), (10-8,10-6,10-1), (4-3,4-3,4-1), (8-0,8-7,8-5)$, or $(7,3,2), (2,4,9), (1,1,4), (8,1,3)$. We make sure the $f_i$ triples are also digits (between 0 and 9).
3. Concatenate corresponding digits of each triple $e_i, f_i$ to form our sum in (1):

$\displaystyle 1830 + 5637 + 6105 = 7218 + 3411 + 2943$

At last we come to the meaning of the title of this post. Integer equations such as

$\displaystyle a + b + c = u + v + w$

$\displaystyle a^2 + b^2 + c^2 = u^2 + v^2 + w^2$

are known as (2,3)-multigrade equations. The 2 represents the highest power and the 3 represents the number of terms in each side of the equation.

The following are known about the solutions to such equations.

1. If $(a,b,c,u,v,w)$ is a solution, so are $(a+k, b+k, c+k, u+k, v+k, w+k)$ and $(ak, bk, ck, dk, uk, vk, wk)$ for any integer $k$.
2. The general solution has the form $(a,b,c,u,v,w)=(g + e,h+e,m+n+e,m+e,n+e,g+h+e)$ where $g,h,e$ are integers satisfying $gh = mn$.
3. Another form for the general solution is $(a,b,c,u,v,w) = (wy+xz+e, wz+e,xy+e,wz+xy+e,xz+e,wy+e)$, where $w,x,y,z,e$ are integers.

Note that 1 is straightforward to show while 2 follows from 3 by setting $g = wz, h = xy, m = xz, n = wy$. We will show 3 at the end of this post.

Not all solutions have the property $a+u =b+v = c+w$ (even after permuting $u,v,w$. This was only needed above to ensure that when concatenating two or more digits the squared equation work (i.e. so that the cross-terms equated). For example, $5^2 + 0^2 + 8^2 = 2^2 + 9^2 + 2^2$ and $8^2 + 6^2 + 1^2 = 2^2 + 4^2 + 9^2$ are valid pairs of (2-3)-multigrade equations but after concatenating to form 2-digit numbers, $58^2 + 06^2 + 82^2 < 22^2 + 94^2 + 82^2$. The solution $(a,b,c,u,v,w)=(5,0,8,2,9,2)$ does not have the $a+u =b+v = c+w$ property (the sum $a + u + b + v + c + w$ would need to be a multiple of 3 for this to be possible).

#### Appearance of (2,3)-multigrade solutions in magic squares

Interestingly, if you take any 3×3 magic square (integers in a 3 by 3 array so that the numbers in each row, column and diagonal have the same sum), the first and third rows or columns form solutions to (2,3)-multigrade equations. As an example, take the following instance (copied from here).

The first and third rows give $2^2 + 7^2 + 6^2 = 4^2 + 3^2 + 8^2 = 89$ while the first and third columns give $2^2 + 9^2 + 4^2 = 6^2 + 1^2 + 8^2 = 101$. Furthermore by concatenation forwards and backwards we can find many more (2,3)-multigrade equations:

$\displaystyle 276^2 + 951^2 + 438^2 = 672^2 + 159^2 + 834^2$ (rows)

$\displaystyle 294^2 + 753^2 + 618^2 = 492^2 + 357^2 + 816^2$ (columns)

$\displaystyle 258^2 + 714^2 + 693^2 = 852^2 + 417^2 + 396^2$ (diagonals)

$\displaystyle 213^2 + 798^2 + 654^2 = 312^2 + 897^2 + 456^2$ (counter-diagonals)

$\displaystyle 258^2 + 936^2 + 471^2 = 852^2 + 639^2 + 174^2$ (diagonals)

$\displaystyle 231^2 + 978^2 + 456^2 = 132^2 + 879^2 + 654^2$ (counter-diagonals)

All the above hold also when the power indices of 2 are replaced with 1. These six relations (first discovered by Dr Irving Joshua Matrix according to [1]) all follow from the single-digit relations using similar reasoning from before.

The single-digit relations can be verified by using the general form of a 3×3 magic square:

 $p + q + 2r$ $p$ $p + 2q + r$ $p + 2q$ $p+q +r$ $p + 2r$ $p + r$ $p + 2q + 2r$ $p + q$

#### General solution to (2,3)-multigrade equations

Finally for completeness we shall derive the general solution of the equation:

$\displaystyle a_1 + a_2 + a_3 = b_1 + b_2 + b_3$

$\displaystyle a_1^2 + a_2^2 + a_3^2 = b_1^2 + b_2^2 + b_3^2$

We use the procedure outlined in [2]. Take $S = a_1 + a_2 + a_3, 3A_i = a_i + S, 3B_i = b_i + S$. Substituting gives

$\displaystyle \sum_{i=1}^3 A_i = \sum_{i=1}^3 B_i = 0, \sum_{i=1}^3 A_i^2 = \sum_{i=1}^3 B_i^2$

This is slightly easier to work with since we can reduce this to an equation of fewer variables. Since $A_3 = -(A_1 + A_2)$,

$\displaystyle \sum_{i=1}^3 A_i^2 = A_1^2 + A_2^2 + (A_1+A_2)^2 = 2(A_1^2 + A_2^2 + A_1A_2)$

and so

$\displaystyle A_1^2 + A_2^2 + A_1A_2 = B_1^2 + B_2^2 + B_1B_2 \quad \quad (2)$

To proceed we use the fact that $x^2 + xy + y^2$ be factorised in the complex numbers as $(x + y(1 + \sqrt{3}i)/2)(x + y(1 - \sqrt{3}i)/2)$.

We use unique factorisation in the quadratic ring $\mathbb{Z}[(1+\sqrt{3}i)/2]$ (the Eisenstein integers). We can write numbers in this ring uniquely in the  form $z = a + b \omega$, where $a,b$ are integers and $\omega = (1+\sqrt{3}i)/2$, one of the cube roots of -1. Visualize Eisenstein integers as elements of an equilateral triangular lattice in the complex plane (as opposed to Gaussian integers which are elements of a square lattice).

Assume that $A_i \neq B_i$ in (2). Then letting both sides of this equation equal $N$, equation (2) represents a factorisation of $N$ in two ways.

$\displaystyle N = (A_1 + A_2\omega) (A_1 + A_2\bar{\omega}) = (B_1 +B_2\omega) (B_1 + B_2\bar{\omega})$

As factorisation in our ring of Eisenstein integers is unique, this equation must imply that one of the terms, say $(A_1 + A_2 \omega)$, must be the product of two non-unit terms, say $(a + b\omega)(c + d\omega)$ where $a,b,c,d$ are integers. Then pairing up products with their conjugates, the factorisation of (2) can be written in the following two ways.

$\displaystyle N = (a + b\omega)(c + d\omega)(a + b\bar{\omega})(c + d\bar{\omega}) = (a + b\omega)(c + d\bar{\omega})(a + b\bar{\omega})(c + d\omega)$

In other words, $A_1 + A_2 \omega = (a + b\omega)(c + d\omega)$ and $B_1 + B_2 \omega = (a + b\omega)(c + d\bar{\omega})$, leading to

$\begin{array}{lcl}A_1 + A_2 \omega &=& ac + (ad + bc)\omega + bd\omega^2\\ &=& ac + (ad + bc)\omega + bd(\omega-1)\\ &=& (ac-bd) + (ad + bc + bd)\omega\end{array}$

$\begin{array}{lcl}B_1 + B_2 \omega &=& ac + ad\bar{\omega} + bc \omega + bd \omega \bar{\omega}\\ &=& (ac + bd) + ad (1 - \omega) + bc \omega\\ &=& (ac + bd + ad) + (bc - ad)\omega\end{array}$

Matching coefficients then gives us the general solution $A_1 = ac-bd, A_2 = ad + bc + bd, B_1 = ac + bd + ad, B_2 = bc - ad$. To reach the $(wy+xz+e, wz+e,xy+e,wz+xy+e,xz+e,wy+e)$ form given above, we first note that among the four integers $a,b,c,d$, two of them, say $a$ and $b$ are the same modulo 3. We then let $a = y + z, b = y - 2z, c =x, d = w - x$ and find that

$\begin{array}{lcl}ac-bd &=& (y+z)x-(y-2z)(w-x)\\&=& xy+xz-wy+xy+2wz-2xz\\&=& 2xy - xz - wy + 2wz\\&=& 3(wz+xy) - (w+x)(y+z)\end{array}$

$\begin{array}{lcl}ad + bc + bd &=& (a+b)(c+d)-ac\\&=& (2y-z)w - x(y+z)\\&=& 2wy-wz-xy-xz\\&=& 3wy - (w+x)(y+z)\end{array}$

$\begin{array}{lcl}ac + bd + ad &=& (a+b)(c+d)-bc\\&=& (2y-z)w - x(y-2z)\\&=& 2wy-wz-xy+2xz\\&=& 3(xz+wy) - (w+x)(y+z)\end{array}$

$\begin{array}{lcl} bc - ad & = & (y - 2z)x - (y+z)(w-x)\\& = & wy - txz - wy - wz + xy + xz\\ & = & 2xy - xz - wy - wz \\ &=& 3xy - (w+x)(y+z)\end{array}$

Hence after adding $S = (w+x)(y+z)$ and dividing by 3 (if all the numbers are divisible by 3), we arrive at the form above. Finally note that adding any constant to each entry also produces a valid solution.

By computational search, we find the single digit non-trivial solutions to $\displaystyle a_1 + a_2 + a_3 = b_1 + b_2 + b_3$ and $\displaystyle a_1^2 + a_2^2 + a_3^2 = b_1^2 + b_2^2 + b_3^2$ (up to permutations) are given by the following 16 forms.

$\displaystyle (0, 3, 3, 1, 1, 4) \quad (0, 4, 5, 1, 2, 6) \quad (0,5,7,1,3,8) \quad (0,5,8,2,2,9)$

$\displaystyle (0,6,6,2,2,8) \quad (0, 7,7,1,4,9) \quad (1,4,4,2,2,5) \quad (1,5,6,2,3,7)$

$\displaystyle (1,6,8,2,4,9) \quad (1,7,7,3,3,9) \quad (2,5,5,3,3,6) \quad (2,6,7,3,4,8)$

$\displaystyle (3,6,6,4,4,7) \quad (3,7,8,4,5,9) \quad (4,7,7,5,5,8) \quad (5,8,8,6,6,9)$

Most of these can be concatenated to form multi-digit (2,3)-multigrade equations such as (1). The only exceptions are those which do not have sum a multiple of 3:  $(0,5,8,2,2,9)$ and $(0,7,7,1,4,9)$. More information about (2,3)-multigrade equations can be found in [3].

#### References

[1] A. Benjamin and K. Yasuda, “Magic Squares Indeed!”, Amer. Math. Monthly, Feb. 1999.

[2] Dickson, History of the theory of numbers, available at: http://archive.org/details/historyoftheoryo02dickuoft

[3] T. Piezas, Sum / Sums of three squares – A Collection of Algebraic Identities: https://sites.google.com/site/tpiezas/004

## August 12, 2012

### The race to most ODIs played

Filed under: cricket,sport — ckrao @ 3:50 am

I noticed that in the recent one-day international series played between Sri Lanka and India, Mahela Jayawardene entered third place for most ODIs played. His tally of 382 only trails Sanath Jayasuriya (445) and Sachin Tendulkar (463). He is still 35 and playing well enough to have a few good years left in him. It will be tough but he is the main contender to break Tendulkar’s record.

The above graph compares the number of games played versus age and it can be seen that the gap is closing, currently down to 35 games for Jayawardene’s present age. At one stage this gap was as many as 93 games for the same age (around 28).

Jayawardene would need to keep up his current activity to bridge the gap sufficiently. Of course all depends on how long the both of them continue to play.

(data from espncricinfo.com)

Create a free website or blog at WordPress.com.