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Plimpton 322 How p and q were Chosen
[edit]In an attempt at clarity, let me first state my belief that there are at least three iterations of Plimpton 322. Working backwards:
3. The clay tablet, a student Find-Fix Exam, sexagesimal with 6 mistakes. 2. The checksheet for the exam, sexagesimal also used for classroom examples. 1. The original table, sexagesimal before Row11 and Row15 were changed.
There is a Plimpton 322 article by Roger C. Alperin that I will reference for my solution to how p and q were chosen to generate Plimpton 322:
http://www.math.sjsu.edu/~alperin/Plimpton322.pdf
On page 1/2 of the pdf is the familiar sexagesimal (corrected) checksheet. The column designated w is a ratio that is often referred to as p/q. As a decimal fraction, it runs down in fifteen steps from 2.4 for Row01 to 1.7778 for Row15; and for Row08 p/q is 2.1333; so it appears to be almost linear. Let's find the p/q for 45deg and 30deg. And for 45deg p/q is 2.4142 and for 30deg p/q is 1.7321.
With a computer algorithm, the p,q pairs from 2,1 to 125,124 can be used to create right triangles between 0deg and 90deg, the possible universe of the article; and there are 3906. Of which, coprime pairs number 3195. Now let's limit p/q to be between 2.4142 and 1.7321, the portion of interest, and there are 518. Each coprime pair produces a pair of primitive Positive Integer Right Triangles, PIRTs. (One of each will be shown below). Sieve out those p/q's with (p-q) equal to a prime number (1 is prime, 2 is not), and get 296. The PIRTs of the original table have (a^2/b^2) exact to 8 sexagesimal fractional places or less. Since the even leg (b) of each PIRT is equal to (2*p*q), let's sieve out those p/q's that meet a 60^8/(2*p*q)^2=int(60^8/(2*p*q)^2) criteria. We are left with 17. Looking for the original table, before Row11 and Row15 were changed to produce the checksheet, we find the 15.
It should be noted that if we use the format proposed in Alperin's article, for Row01 that 60^8/(2*p*q)^2 becomes (2^16*3^8*5^8)/(2^6*3^2*5^2)=(2^10*3^6*5^6), easily seen as an integer result,(16*729*10^6). A more complete discussion of this exactness is to be found on page 2/7 of PLIMPTON 322 Take Another Look by T. L. O'Donnell at:
http://members.localnet.com/~tomo/
In an article published in 2002:
http://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Robson105-120.pdf Words and Pictures: New Light on Plimpton 322 by Eleanor Robson
In the middle of page 3/16 of the pdf, Robson comments, "The quest has been to find how p and q were chosen". The quest is over! Well, maybe not quite.
There is a branch of mathematics called Number Theory. And it has a section on Positive Integer Right Triangles, PIRTs. It can start out with combining construction of dual PIRTs by p,q pairs with the odd&even business, they seem to go together. They are after all, both complementary pairs. And it is also a good time to introduce Plimpton 322.
The four a,b possibilities: | Rules for dual primitive PIRT construction by p,q pairs:
1. a is odd, b is even. p and q are positive integers, one even, one odd. 2. a is even, b is odd. p > q > 0; p,q are coprime, only common factor is l. 3. Both are even. 1st outcome: a=p^2-q^2, b=2pq, c=p^2+q^2. Possibility 1. 4. Both are odd. 2nd outcome: a=2pq, b=p^2-q^2, c=p^2+q^2. Possibility 2.
For 1. The original Plimpton 322 table provides 15 primitive PIRTs; a is odd and b is even; in the 45deg to 30deg range. In these cases a=(p^2-q^2), b=(2*p*q), and c=(p^2+q^2); for Row11: a=3, b=4, and c=5. How about when a is odd, b is even, and the PIRT is not primitive. Multiply all sides of Row11 by an odd integer, 15 will do. (Teacher's age, perhaps). And: a=45, b=60, and c=75, is not primitive and becomes Row11 of the checksheet.
For 2. But what about when a is even and b is odd. We should look at the 2nd outcome in the complementary range, 45deg to 60deg. And for 60deg p/q is 3.7321 and for 45deg p/q is 2.4142. So, we look there at the same p,q pairs from 2,1 to 125,124 and still find 3906, of which coprimes are still 3195, but those of interest are 471 (slightly lower). And only 1 has 60^8/(p^2-q^2)^2 an integer at p/q=7/2=3.5. In this case the even leg a=(2*p*q), b=(p^2-q^2), and c=(p^2+q^2); and: a=28, b=45, and c=53 is primitive.
For 3. Multiply all sides of this PIRT by an even integer, 2 will do, and get a=56, b=90, and c=106. This is not primitive and becomes Row15 of the checksheet.
For 4. 3800 years later and still looking. (Not possible and I have a proof).
Note: For 3. Row15 of the original table; a=175, b=288, c=337, c^2/b^2=1.3692 could be replaced by a=56, b=90, c=106, c^2/b^2=1.3872 on the checksheet because of the c^2/b^2 match-ups.
Now we're done!
Thomas L. O'Donnell 03/27/2017 .336sunny (talk) 04:50, 25 April 2017 (UTC)
Plimpton 322 from Reduced Triples.
[edit]This topic will continue the emphasis that the tablet was created from primitive right triangles.
Plimpton 322 (corrected) is a table of 15 Positive integer Right Triangles, PIRTs. 14/15 are in the 45deg to 30deg range with 2.4142>(p/q)>1.7321.
13/15 are primitive, that is irreducible. And since a is odd and b is even for the 13, this set of "reduced triples" should be used:
a=((p/q)-(q/p))/2 b=1 c=((p/q)+(q/p))/2 with p/q=9/4, a=((9/4)-(4/9))/2 b=1 c=((9/4)-(4/9))/2 and solving: a=65/72 b=1 c=97/72 multiply all sides of this "reduced triple" by 72 to get Row05: a=65 b=72 c=97 a primitive PIRT and (b) is a regular sexagesimal number.
So this is the set to use when the PIRT is primitive and a is odd, b is even.
A regular sexagesimal is a whole number that can be put in the form of (2^i)(3^j)(5^k); where i, j and k can be any integer including 0. To put it another way; it can not be a prime number of 7 or above, nor a multiple.
And we got rid of those pesky squares in the complimentary pairs:
a=p^2-q^2 b=2*p*q c=p^2+q^2 a is odd, b is even a=2*p*q b=p^2-q^2 c=p^2+q^2 a is even, b is odd
For Row11 use p/q=2/1 and get
a=3 b=4 c=5 multiply thru by the common factor, 15. a=45 b=60 c=75 not a primitive and (b) is a "regular" number.
From Row15 we get a sample of what the Teacher is teaching. p/q=7/2. And we need to realize that this p/q is in the complementary range, 45deg to 60deg, 2.4142<(p/q)<3.7321, and therefore we want the alternate set (which would be in the 45deg to 30deg range). Using p/q=7/2, our "reduced triple" alternate set becomes:
a=1 b=((p/q)-(q/p))/2 c=((p/q)+(q/p))/2 with p/q=7/2, a=1 b=((7/2)-(2/7))/2 c=((7/2)+(2/7))/2 and solving: a=1 b=45/28 c=53/28 and multiplying all sides of this "reduced triple" by 28: a=28 b=45 c=53 a primitive PIRT, multiply by the common factor 2, and get Row15: a=56 b=90 c=106 not a primitive
So this is the set to use when the PIRT is primitive and a is even, b is odd. And the Teacher would have pointed out that (b) should be a "regular" number, and it is.
This last paragraph has been appended to the Plimpton 322 Talk page.
http://en.wiki.x.io/wiki/Talk:Plimpton_322#Problems_with_the_"problem_list"_theory
A discussion of Plimpton 322 is like an introduction to Number Theory. Integer math (arithmetic), odd & even, prime numbers, elementary algebra, and positive integer right triangles are all included.
http://en.wiki.x.io/wiki/Number_theory#Dawn_of_arithmetic
It's hard to recognize that the equation that's shown there is for a reduced triple. Who would know that x=(p/q)? And that primitive triples exist in complementary pairs?
PIRT generation with p and q between 1 and 125
[edit]Go to http://members.localnet.com/~tomo TABLE1 Proposed Missing Part for (d/l)^2 brackets. TABLE1 was filled in the order PIRTs were found. X15 was skipped because the bracket was already filled. Search was ended when #04 was found and before PIRT's X12 and X16 were reached.
Results of primitive PIRT generation by p,q's between 1 and 125 with (l) a regular sexagesimal
#11 (d/l)^2=1.562 p=2 q=1 p/q=2 s=3 l=4 d=5 --- (d/l)^2=1.174 p=3 q=2 p/q=1.5 s=5 l=12 d=13 --- (d/l)^2=1.284 p=4 q=1 p/q=4 s=8 l=15 d=17 --- (d/l)^2=1.085 p=4 q=3 p/q=1.33 s=7 l=24 d=25 --- (d/l)^2=1.051 p=5 q=4 p/q=1.25 s=9 l=40 d=41 --- (d/l)^2=1.034 p=6 q=5 p/q=1.2 s=11 l=60 d=61 #15 (d/l)^2=1.387 p=7 q=2 p/q=3.5 s=28 l=45 d=53 --- (d/l)^2=1.238 p=8 q=5 p/q=1.6 s=39 l=80 d=89 #05 (d/l)^2=1.815 p=9 q=4 p/q=2.25 s=65 l=72 d=97 --- (d/l)^2=1.014 p=9 q=8 p/q=1.12 s=17 l=144 d=145 --- (d/l)^2=1.011 p=10 q=9 p/q=1.11 s=19 l=180 d=181 #01 (d/l)^2=1.983 p=12 q=5 p/q=2.4 s=119 l=120 d=169 #13 (d/l)^2=1.45 p=15 q=8 p/q=1.88 s=161 l=240 d=289 X15 (d/l)^2=1.369 p=16 q=9 p/q=1.78 s=175 l=288 d=337 --- (d/l)^2=1.004 p=16 q=15 p/q=1.07 s=31 l=480 d=481 #06 (d/l)^2=1.785 p=20 q=9 p/q=2.22 s=319 l=360 d=481 #09 (d/l)^2=1.643 p=25 q=12 p/q=2.08 s=481 l=600 d=769 --- (d/l)^2=1.213 p=25 q=16 p/q=1.56 s=369 l=800 d=881 --- (d/l)^2=1.112 p=25 q=18 p/q=1.39 s=301 l=900 d=949 --- (d/l)^2=1.002 p=25 q=24 p/q=1.04 s=49 l=1200 d=1201 --- (d/l)^2=1.006 p=26 q=1 p/q=26 s=52 l=675 d=677 --- (d/l)^2=1.3 p=27 q=16 p/q=1.69 s=473 l=864 d=985 --- (d/l)^2=1.093 p=27 q=20 p/q=1.35 s=329 l=1080 d=1129 #08 (d/l)^2=1.693 p=32 q=15 p/q=2.13 s=799 l=960 d=1249 --- (d/l)^2=1.062 p=32 q=25 p/q=1.28 s=399 l=1600 d=1649 --- (d/l)^2=1.029 p=32 q=27 p/q=1.19 s=295 l=1728 d=1753 --- (d/l)^2=1.139 p=36 q=25 p/q=1.44 s=671 l=1800 d=1921 --- (d/l)^2=1.163 p=40 q=27 p/q=1.48 s=871 l=2160 d=2329 --- (d/l)^2=1.121 p=45 q=32 p/q=1.41 s=1001 l=2880 d=3049 #12 (d/l)^2=1.489 p=48 q=25 p/q=1.92 s=1679 l=2400 d=2929 #14 (d/l)^2=1.43 p=50 q=27 p/q=1.85 s=1771 l=2700 d=3229 #07 (d/l)^2=1.72 p=54 q=25 p/q=2.16 s=2291 l=2700 d=3541 #02 (d/l)^2=1.949 p=64 q=27 p/q=2.37 s=3367 l=3456 d=4825 --- (d/l)^2=1.129 p=64 q=45 p/q=1.42 s=2071 l=5760 d=6121 #03 (d/l)^2=1.919 p=75 q=32 p/q=2.34 s=4601 l=4800 d=6649 --- (d/l)^2=1.025 p=75 q=64 p/q=1.17 s=1529 l=9600 d=9721 #10 (d/l)^2=1.586 p=81 q=40 p/q=2.02 s=4961 l=6480 d=8161 --- (d/l)^2=1.251 p=81 q=50 p/q=1.62 s=4061 l=8100 d=9061 --- (d/l)^2=1.057 p=81 q=64 p/q=1.27 s=2465 l=10368 d=10657 --- (d/l)^2=1 p=81 q=80 p/q=1.01 s=161 l=12960 d=12961 --- (d/l)^2=1.045 p=100 q=81 p/q=1.23 s=3439 l=16200 d=16561 --- (d/l)^2=1.2 p=103 q=22 p/q=4.68 s=4532 l=10125 d=11093 #04 (d/l)^2=1.886 p=125 q=54 p/q=2.31 s=12709 l=13500 d=18541 X12 (d/l)^2=1.519 p=125 q=64 p/q=1.95 s=11529 l=16000 d=19721 X16 (d/l)^2=1.336 p=125 q=72 p/q=1.74 s=10441 l=18000 d=20809 --- (d/l)^2=1.071 p=125 q=96 p/q=1.3 s=6409 l=24000 d=24841 --- (d/l)^2=1.022 p=125 q=108 p/q=1.16 s=3961 l=27000 d=27289