# The Battle Of Hastings

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• July 26th 2007, 07:09 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
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.

Let me see if I've got this right then:

$x^2 - 2y^2 = 1$

$\sqrt{2} = [1; 2, 2, 2, ~ ... ~ ]$

Thus $l = 1$ and l is odd.

So I have
$C_{2l - 1} = C_{2 \cdot 1 - 1} = C_1 = 1 + \frac{1}{2} = \frac{3}{2} = \frac{x}{y}$

Thus $y = \frac{2}{3}x$.

So
$x^2 - 2\left ( \frac{2}{3} x \right )^2 = 1 ==> x = 3$. Thus $y = 2$. Which is a correct solution.

-Dan
• July 26th 2007, 08:12 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Let me see if I've got this right then:

$x^2 - 2y^2 = 1$

$\sqrt{2} = [1; 2, 2, 2, ~ ... ~ ]$

Thus $l = 1$ and l is odd.

So I have
$C_{2l - 1} = C_{2 \cdot 1 - 1} = C_1 = 1 + \frac{1}{2} = \frac{3}{2} = \frac{x}{y}$

Thus $y = \frac{2}{3}x$.

So
$x^2 - 2\left ( \frac{2}{3} x \right )^2 = 1 ==> x = 3$. Thus $y = 2$. Which is a correct solution.

-Dan

You solution is correct but the procedure is not.

$\frac{3}{2}$ means chose $x=\mbox{ numerator }=3$ and $y=\mbox{ deminator }=2$.
• July 26th 2007, 08:31 AM
galactus
Quote:

Harold must have been quite a leader!

I reckon not. He got shot in the eye at Hastings and the Normans won.
• July 26th 2007, 08:41 PM
Soroban
Hello, Galactus!

Quote:

I reckon not. He got shot in the eye at Hastings and the Normans won.
I was referring to the size of his army.

Since $y = 226,153,980$, the first formation had: . ${\color{blue}3.18 \times10^{18}}\,\text{ soldiers.}$

And the Normans won?
They must have drafted Klingons, Vogans,and Trafalmadorians.

• July 27th 2007, 02:34 AM
galactus
I know what you meant, Soroban. Despite the astronomical size of his army, he still lost. Though, I am pretty sure, his army was not actually that large. Over 3 quintillion in the first formation. You would think such an army would be invincible.;)

I believe the realsitic size of the Saxon army is thought to be have been about 8,000 or so.

Just relating history.
• July 27th 2007, 04:07 AM
henryzz
Continued Fraction Calculator

$
x^2 - 61y^2 = 1
$

$
\sqrt{61} = [7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 ~ ... ~ ]
$

Thus $l = 11$and l is odd

So I have
[tex] C_{2l - 1} = C_{2 \cdot 11 - 1} = C_{21} = 7 + \frac{1}{1+\frac{1}{4+\frac{1}{3+\frac{1}{1+\frac{1}{2+\frac{1}{2+\frac{1}{1+\frac{1}{3+\frac{1}{4+\frac{1}{1+\frac{1}{14+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{3+ \frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{2+\frac{ 1}{1+\frac{1}{2+\frac{1}{2+\frac{1}{1+\frac{1}{1+\ frac{1}{3+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1 }{1}}}}}}}}}}}}}}}}}}}}}}}}}}}[/math]
$= \frac{2713847760}{183241189}$
so
$
x = 2713847760
$

$
y = 183241189
$

there is something wrong with the result
all the maths was done by the site above
can u point out what i did wrong

edit: i have corrected 1 mistake are there any others
edit: whats making it show the code
• July 27th 2007, 04:53 AM
topsquark
Quote:

Originally Posted by henryzz
Continued Fraction Calculator

$
x^2 - 61y^2 = 1
$

$
\sqrt{61} = [7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 ~ ... ~ ]
$

Thus $l = 7$and l is odd

How are you getting l = 7? I'm counting 11 digits in the repeating part.

-Dan
• July 27th 2007, 05:25 AM
henryzz
thanks will edit
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