# Solution

(related to Problem: The Battle of Hastings)

Any number (not itself a square number) may be multiplied by a square that will give a product $1$ less than another square. The given number must not itself be a square because a square multiplied by a square produces a square, and no square plus $1$ can be a square. My remarks throughout must be understood to apply to whole numbers, because fractional soldiers are not of much use in war.

Now, of all the numbers from $2$ to $99$ inclusive, $61$ happens to be the most awkward one to work, and the lowest possible answer to our puzzle is that Harold's army consisted of $3,119,882,982,860,264,400$ men. That is, there would be $51,145,622,669,840,400$ men (the square of $226,153,980$) in each of the sixty-one squares. Add one man (Harold), and they could then form one large square with $1,766,319,049$ men on every side. The general problem, of which this is a particular case, is known as the "Pell's Equation" — apparently because Pell neither first propounded the question nor first solved it! It was issued as a challenge by Pierre de Fermat to the English mathematicians of his day. It is readily solved by the use of continued fractions.

Next to $61,$ the most difficult number under $100$ is $97,$ where $97 \times 6,377,352^2 + 1 = n$, where $n$ is a a square.

The reason why I assumed that there must be something wrong with the figures in the chronicle is that we can confidently say that Harold's army did not contain over three trillion men! If this army (not to mention the Normans) had had the whole surface of the earth (sea included) on which to encamp, each man would have had slightly more than a quarter of a square inch of space in which to move about! Put another way: Allowing one square foot of standing-room per man, each small square would have required all the space allowed by a globe three times the diameter of the earth.

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### References

#### Project Gutenberg

1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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