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| The Garden of Pythagoras (Posted on 2010-08-03) |
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The god Zeus commanded the Sybarites to furnish his temple with a large piece of land on which to construct a garden or precinct.
Not wishing to defy the god, but reluctant to part with so much land, the Sybarites made the donation subject to conditions which they believed could not be fulfilled. They required that the garden be laid out with an open central square, abutted by the hypotenuses of 4 right triangular groves,
such that:
1. All dimensions of the square and triangles must be measurable in whole numbers of cubits;
2. No two of the outer sides of the triangles should be of the same length;
3. No two sides of any triangle should have a common divisor.
4. The whole should be of the minimum size permitted by the foregoing requirements.
The priests of Zeus turned to Pythagoras for assistance. To the consternation of the Sybarites, Pythagoras not only immediately produced a plan compliant with these specifications, but into the bargain made proposals for a grand estate, laid out in like manner, but with an octagonal centerpiece!
What was the length (in cubits) of the sides of the central square in the original plan?
Bonus question: approximately how many times larger than the original would the surface area of the larger project proposed by Pythagoras have been?
A cubit is about 50cm.
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Submitted by broll
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Pythagoras’s Garden
Main Question.
1. The solution has to be the smallest possible, where the shorter sides (call them the x and y sides) have no common factor. Call such a solution a ‘primitive’ solution.
2. The Pythagorean formula is y = u^2-v^2, x = 2uv, z = u^2 + v^2. What can we say about z?
3. We know that one of the short sides has to be even (2uv). It follows trivially that both the other sides must be odd. So the z side is odd.
4. We know or it can trivially be shown (e.g. by drawing a square and dividing the sides as near the centre of the sides as possible) that all even squares are of the form 4k and all odd squares are of the form 4k+1. An even square plus an odd square must produce a result in the same form, 4k+1. Z^2 must be of this form. (This doesn’t mean -yet- that z also has to be; e.g. 3^2 is 9, which is of the form 4k+1)
5. We know or it can trivially be shown that all odd numbers - specifically, odd primes - are either of the form 4k+1 or 4k+3.
6. If z is to be the sum of 2 squares in more than 1 way, then z is composite, since by descent, a prime of form 4k+3 is never the sum of 2 squares, while a prime of form 4k+1 is the sum of 2 squares in exactly one way.
7. We know by the fundamental theorem of arithmetic that every non-prime number can be reduced to a unique product of prime factors; prime numbers or their powers. We can exclude 2 for obvious reasons. All primes other than 2 are odd.
8. Again by descent, both z and every prime factor of z must themselves be of the same form, 4k+1. (A power of such a prime also qualifies under the fundamental theorem, but counts only once – no prime power is itself prime).
9. We can therefore say the following things about z:
a. z is odd (3 above)
b. z^2 is of the form 4k+1 (4 above)
c. z is a compound number (6 above)
d. the factors of z are themselves primes of the form 4k+1, or their powers (8 above)
e. since every such representation is unique (7 above), and since powers of such numbers are larger than the numbers of which they are powers, the least such representation of z for any given number of ways will be of the form 5*13*17… where each factor of z is the next higher prime of the form 4k+1.
12. Consider such representations: 5 by itself gives 3,4,5; which could fairly be described as the least primitive solution with exactly 1 x and y. 5*13 is 65, which has the x,y pairs 8,1 and 7,4. By straightforward combination, each time a new prime is added, the number of ways is doubled: z = 5*13*17 gives the result z = 1105, which hypotenuse is common to the primitive solutions 47,1104,1105; 744,817,1105; 576,943,1105; and 264,1073,1105.
13. Since 1105 is the smallest possible number with 3 different prime components of the form 4k+1, that is the solution to the problem.
Supplemental Question.
There is an exact answer and a rough answer!
It’s not hard to see that the next solution will be 5*13*17*29 = 32045. Obviously this is 29^2 = 841 times the size of 1105. If we allow for the fact that there are twice as many triangles, and use 2* 841 or 1682 as multiplier for the area of the triangles, the result is not far out. The area of the octagon is 2*(1+sqrt2)* 32045^2, for which we can write (1+sqrt2)*2*(1105*29)^2; so the proportion is (1+sqrt2)*2*(1105*29)^2/(1105^2): (1+sqrt2)*1682 or 4061:1; remarkably, 4061 is also almost exactly 29^3/6. These figures give a good approximation to the size of the large project:
If we calculate the area of the ‘small’ garden we get
1221025 Plaza
25944 small triangles
303924 small triangles
271584 small triangles
141636 small triangles
743088
1964113 Total
So the bigger estate will be of approximate size: 1221025*4061, 743088*1682 = 4958582525+ 1249874016, or 6,208,456,541 square cubits.
This corresponds quite well to the actual value of 6,205,484,743.4 (the error is less than .05%)
So the larger project is approximately 3160 times ( or almost exactly 29*109 times) the smaller one.
Assuming a cubit is about 50cm, the smaller garden would be almost exactly 0.5sqkm, while the larger estate would be 1551sqkm ; about twice the size of Singapore, or 1/3 larger than Hong Kong. Ancient writers suggest that the area within the walls of the city of Sybaris was around 8 square kilometers; Pythagoras’s second solution must therefore have come as a nasty shock to the Sybarites!
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