THE BATTLE OF HASTINGS

[henryzz]

THE 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.


i know this is bascically pell's equation but all the pell's equation solvers on the net use decimals this puzzle needs a integer answer
is there a program on the net which will give an ineger result to a pell's equation

[retina]

Unless you count 0 as an answer then there are no solutions that can be accommodated up the size of the entire population of the Earth. Beyond this I have not tested.

[S485122]

start with a square = n^2
n^2=n^2-1+1
n^2=(n+1)*(n-1)+1
we also have 61*m+1=n^2, m an integer
61*m=(n-1)*(n+1)

61 being prime the smallest solution is n+1=61 : 61*59+1=3600=60^2

the next solution is n-1=61 giving 61*63+1=3844=62^2

Jacob

[wblipp]

Originally Posted by henryzz

is there a program on the net

One of Dario Alpern's java applets can be used to solve 61x²-y²+1, which is the same thing.

http://www.alpertron.com.ar/QUAD.HTM

Jacob - you have solved 61x+1=y², not 61x²+1=y²

[retina]

Thanks to wblipp's contribution the answer is:

226153980^2*61+1=1766319049^2

Thus the number of men is 3119882982860264400+1

[henryzz]

that applet is super quick

i wrote a program to solve the equation dx2+1=y2 where i started d at 1 and incremented it
it finds d=61 in about 5 minutes the first it coulnt find was 109 which this does in no time

[retina]

Originally Posted by henryzz

that applet is super quick

It is not a brute force program. It uses algebra and a few algorithms. Done very nicely BTW.

Originally Posted by henryzz

... it finds d=61 in about 5 minutes the first it coulnt find was 109 ...

Try 8191 or 65521.

[henryzz]

super fast