# Math Help - The Battle Of Hastings

1. ## The Battle Of Hastings

129THE BATTLE OF HASTINGS
All historians know that there is a great deal of mystery and uncertainty concerning the details of the ever-memorable battle on that fatal day, October 14, 1066. My puzzle deals with a curious passage in an ancient monkish chronicle that may never receive the attention that it deserves, and if I am unable to vouch for the authenticity of the document it will none the less serve to furnish us with a problem that can hardly fail to interest those of my readers who have arithmetical predilections. Here is the passage in question.
"The men of Harold stood well together, as their wont was, and formed sixty and one squares, with a like number of men in every square thereof, and woe to the hardy Norman who ventured to enter their redoubts; for a single blow of a Saxon war-hatchet would break his lance and cut through his coat of mail.... When Harold threw himself into the fray the Saxons were one mighty square of men, shouting the battle-cries, 'Ut!' 'Olicrosse!' 'Godemitè!'"
Now, I find that all the contemporary authorities agree that the Saxons did actually fight in this solid order. For example, in the "Carmen de Bello Hastingensi," a poem attributed to Guy, Bishop of Amiens, living at the time of the battle, we are told that "the Saxons stood fixed in a dense mass," and Henry of Huntingdon records that "they were like unto a castle, impenetrable to the Normans;" while Robert Wace, a century after, tells us the same thing. So in this respect my newly-discovered chronicle may not be greatly in error. But I have reason to believe that there is something wrong with the actual figures. Let the reader see what he can make of them.
The number of men would be sixty-one times a square number; but when Harold himself joined in the fray they were then able to form one large square. What is the smallest possible number of men there could have been?
In order to make clear to the reader the simplicity of the question, I will give the lowest solutions in the case of 60 and 62, the numbers immediately preceding and following 61. They are 60 × 42 + 1 = 312, and 62 × 82 + 1 = 632. That is, 60 squares of 16 men each would be 960 men, and when Harold joined them they would be 961 in number, and so form a square with 31 men on every side. Similarly in the case of the figures I have given for 62. Now, find the lowest answer for 61.

whats the best way of solving this puzzle
the only way i can find is by guess-work
this takes a computer about 5 seconds to do
can anyone solve this by hand

2. Originally Posted by henryzz
129THE BATTLE OF HASTINGS
All historians know that there is a great deal of mystery and uncertainty concerning the details of the ever-memorable battle on that fatal day, October 14, 1066. My puzzle deals with a curious passage in an ancient monkish chronicle that may never receive the attention that it deserves, and if I am unable to vouch for the authenticity of the document it will none the less serve to furnish us with a problem that can hardly fail to interest those of my readers who have arithmetical predilections. Here is the passage in question.
"The men of Harold stood well together, as their wont was, and formed sixty and one squares, with a like number of men in every square thereof, and woe to the hardy Norman who ventured to enter their redoubts; for a single blow of a Saxon war-hatchet would break his lance and cut through his coat of mail.... When Harold threw himself into the fray the Saxons were one mighty square of men, shouting the battle-cries, 'Ut!' 'Olicrosse!' 'Godemitè!'"
Now, I find that all the contemporary authorities agree that the Saxons did actually fight in this solid order. For example, in the "Carmen de Bello Hastingensi," a poem attributed to Guy, Bishop of Amiens, living at the time of the battle, we are told that "the Saxons stood fixed in a dense mass," and Henry of Huntingdon records that "they were like unto a castle, impenetrable to the Normans;" while Robert Wace, a century after, tells us the same thing. So in this respect my newly-discovered chronicle may not be greatly in error. But I have reason to believe that there is something wrong with the actual figures. Let the reader see what he can make of them.
The number of men would be sixty-one times a square number; but when Harold himself joined in the fray they were then able to form one large square. What is the smallest possible number of men there could have been?
In order to make clear to the reader the simplicity of the question, I will give the lowest solutions in the case of 60 and 62, the numbers immediately preceding and following 61. They are 60 × 42 + 1 = 312, and 62 × 82 + 1 = 632. That is, 60 squares of 16 men each would be 960 men, and when Harold joined them they would be 961 in number, and so form a square with 31 men on every side. Similarly in the case of the figures I have given for 62. Now, find the lowest answer for 61.

whats the best way of solving this puzzle
the only way i can find is by guess-work
this takes a computer about 5 seconds to do
can anyone solve this by hand

$
61n^2+1=m^2
$

is an example of Pell's equation set by Fermat as a challenge problem in the 17th Century.

Google for Pell's equation and you should find an explanation of the method of finding the solutions.

Alternativly ImPerfectHacker will be along later and I'm sure he will be happy to explain how to solve it.

RonL

3. Hello, Henry!

The number of men would be sixty-one times a square number,
but when Harold himself joined in the fray they were then able to form one large square.
What is the smallest possible number of men there could have been?

The equation is: . $61a^2 + 1 \:=\:b^2\quad\Rightarrow\quad b^2 - 61a^2 \:=\:1$ . for integers $a,\,b$.

Do a search for "Pellian equation".

Edit: RonL beat me to it . . .

4. Euler/Lagrange approach to solving Pell's equation is first to seek the continued fraction for $\sqrt{61}$ which is:
$\sqrt{61} = [7;1,4,3,1,2,2,1,3,4,1,14...]$
The continued fraction begins repeating after $14$. So its length is $l=11$. The simpliest solution to this problem is given by $n/m$ where this fraction is the convergent $C_{21}$.

Thus, we need to compute the fraction of $\sqrt{61}$ extended 21 places . This can be easily be done with a computer program. I am just to lazy to make one.

If anybody is interested I can give you the code to how to compute convergents.

5. the puzzle was originally set by archimedes well before pell
does anyone know how he solved it
i meant how to find it by hand

6. Originally Posted by henryzz
the puzzle was originally set by archimedes well before pell
does anyone know how he solved it
i meant how to find it by hand
Well before Fermat, Pell actualy had nothing to do with this problem.

RonL

7. The equation for inaccurately called "Pell's Equation" by Euler. Since everybody loves Euler nobody cared to correct him.

unfortunatly it was never published and was lost
could anyone find a similar way

9. Originally Posted by henryzz
could anyone find a similar way
I told you what you had to do. It involves finding the continued fraction to 21 places. Which is a very long long job by hand. You can make a program that does that for you.

10. Originally Posted by ThePerfectHacker
I told you what you had to do. It involves finding the continued fraction to 21 places. Which is a very long long job by hand. You can make a program that does that for you.
Sorry, but I'm fuzzy. What do you do once you have this? How do you use it? (I've brushed up on my continued fractions in order to understand this post, but I haven't found any methods involving them beyond GCFs and the Euclidean algorithm. Though the theorem that square roots have repeating digits in the continued fraction representation was a fascinating revelation.)

-Dan

11. Originally Posted by topsquark
Sorry, but I'm fuzzy. What do you do once you have this?
Given a contined fraction,
$[a_0;a_1,a_2,...]$
We define,
$C_0=a_0$
$C_1=[a_0;a_1] = a_0+\frac{1}{a_1}$
$C_2=[a_0,a_1,a_2]=a_0+\frac{1}{a_1+\frac{1}{a_2}}$
Those are called the convergents.

Here is an amazing theorem from Diophantine Approximation theory. It shows that continued fractions are the "best" approximators.

Theorem: Let $x$ be an irrational number. If the reduced fraction $p/q$ satisfies:
$\left| x - \frac{p}{q} \right| < \frac{1}{2q^2}$
Then $p/q=C_k$ for some $k$.
Meaning, it is one of the convergents for the continued fraction expansion of $x$.

Theorem: Given the equation $x^2-ny^2=1$ and $(x,y)=(p,q)$ is a positive solution to this diophantine equation then $p/q$ is one of the convergents of $\sqrt{n}$.

Proof: We have $(x-y\sqrt{n})(x+y\sqrt{n})=1$. Thus, $\frac{p}{q} - \sqrt{n} = \frac{1}{q(p+q\sqrt{n})}$
Thus,
$0<\frac{p}{q}-\sqrt{n} < \frac{\sqrt{n}}{q(q\sqrt{n}+q\sqrt{n})} = \frac{\sqrt{n}}{2q^2\sqrt{n}} = \frac{1}{2q^2}$.
By above theorem the result is complete.

This, shows that all the solutions to the Pell equation are found amongst the convergents. There is an algorithm that says how to find these convergents based on the length of the continued fraction. By using that algorithm we find that $C_{21}$ is the most basic solution.

12. Originally Posted by ThePerfectHacker
Thus, we need to compute the fraction of $\sqrt{61}$ extended 21 places .
So far for the general method I'm good. How did you get that you need the $C_{21}$ convergent?

Or would this conversation be simpler if I proposed solving
$x^2 - 2y^2 = 1$
for integer x and y? (That way it would be much easier to compute the convergent since $\sqrt{2} = [1; 2, 2, 2, ~ ... ]$.)

-Dan

13. sorry i dont really understand continued fractions could u give a link to where it is explained on this forum

14. Harold must have been quite a leader!

The least solution for the Pellian, . $x^2 - 61y^2\;=\;1$ .is: . $\begin{Bmatrix} x & = & 1,766,319,049 \\ y & = & 226,153,980\end{Bmatrix}$

15. Originally Posted by topsquark
So far for the general method I'm good. How did you get that you need the $C_{21}$ convergent?

Or would this conversation be simpler if I proposed solving
$x^2 - 2y^2 = 1$
for integer x and y? (That way it would be much easier to compute the convergent since $\sqrt{2} = [1; 2, 2, 2, ~ ... ]$.)

-Dan
Let $l$ be the length of the square root fraction.

If $l$ is even then $C_{l-1}$ is the first solution.

If $l$ is odd then $C_{2l-1}$ is the first solution.

Once you compute the convergent in reduced terms you will get $p/q$. Then choose $x=p\mbox{ and }y=q$ to obtain a solution.

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