PLIMPTON 322 Take Another Look by Thomas L. O'Donnell. Version 01

Here are four tables that show the application and development of a student sexigesimal Find-Fix exercise in a class on Positive Integer Right Triangles. The class could be part of a course in the theory of numbers. And would be appropriate for students who have had algebra and trigonometry. Tables 1and2 cover the use of the

csc^2(45deg)=2, csc^2(30deg)=4/3. 15deg= 2-(4/3)=2/3.

Row spacing=1deg=(2/3)/15=2/45.

Use this spacing to set up 15 brackets from 45deg down to 31deg.

#1 bracket, 45deg, is 2 +&- 1/45=90/45 +&- 1/45 is 91/45 and 89/45.

#15 bracket, 31deg, is 2-(14*2/45)=62/45 +&- 1/45 is 63/45 and 61/45.

Start=91/45=2.02222222222222, and End=61/45=1.35555555555556 Start-End=2/3

Must Lie Between |

2.02222222222222 |

1.97777777777778 |

1.93333333333333 |

1.88888888888889 |

1.84444444444444 |

1.8 |

1.75555555555556 |

1.71111111111111 |

1.66666666666667 |

1.62222222222222 |

1.57777777777778 |

1.53333333333333 |

1.48888888888889 |

1.44444444444444 |

1.4 |

1.35555555555556 |

1+(a^2/b^2) | a | c | # |

1.98340277777778 | 119 | 169 | 1 |

1.94915855208869 | 3367 | 4825 | 2 |

1.91880212673611 | 4601 | 6649 | 3 |

1.88624790672154 | 12709 | 18541 | 4 |

1.81500771604938 | 65 | 97 | 5 |

1.78519290123457 | 319 | 481 | 6 |

1.71998367626886 | 2291 | 3541 | 7 |

1.69270941840278 | 799 | 1249 | 8 |

1.64266944444444 | 481 | 769 | 9 |

1.58612256611035 | 4961 | 8161 | 10 |

1.5625 | 45 | 75 | 11 |

1.48941684027778 | 1679 | 2929 | 12 |

1.45001736111111 | 161 | 289 | 13 |

1.43023882030178 | 1771 | 3229 | 14 |

1.38716049382716 | 56 | 106 | 15 |

b | b^2 | a^2 |

120 | 14400 | 14161 |

3456 | 11943936 | 11336689 |

4800 | 23040000 | 211669201 |

13500 | 182250000 | 161518681 |

72 | 5184 | 4225 |

360 | 129600 | 101761 |

2700 | 7290000 | 5248681 |

960 | 921600 | 638401 |

600 | 360000 | 231361 |

6480 | 41990400 | 24611521 |

60 | 3600 | 2025 |

2400 | 5760000 | 2819041 |

240 | 57600 | 25921 |

2700 | 7290000 | 3136441 |

90 | 8100 | 3136 |

csc^2=1+(a^2/b^2) rounded to 14 decimal places. Corrected entries underlined.

The 15 brackets just barely fit over the 1+(a^2/b^2) values; Row04 is in by 0.0026 at the top and Row12 is in by 0.0005 at the bottom.

Fourteen of the 16 bracket boundries have repeated numerals. It can be seen that the upper for Row10 can be rewrtten as 1+(6/10)+(2/90) or 1+(36/60)+(80/3600)=1+(37/60)+(20/3600), and that in sexigesimal would be**1:37;20**. And the lower as 1+(5/10)+(7/90)=1+(30/60)+(280/3600)=1+(34/60)+(40/3600), and that in sexigesimal would be **1:34;40**

Anybody that thinks making up this test was easy, look at 1+(a^2/b^2), Row10. Get out pencil and paper and divide 24611521 by 41990400 and get a rounded off 0.58612256611034903215973174820911... . After adding the 1, this is equivalent to making the sexigesimal calculation and getting it__exact__ in 8 fractional places, **1:35;10;2;28;27;24;26;40**. Note that it does fit in its bracket as described above.

To get the contribution of**40** in that 8th place in Row10 to the a^2 numerator, divide **40** by 4*10^6, the result from 60^8/b^2, explained below. And it is a miniscule 0.00001 contribution to the 24611521 numerator!

But how would we know beforehand that a^2/b^2 would be eight sexigesimal places or less? One way would be to see if it could have a common denominator with 60^8. To determine that, divide b^2 into 60^8, and then multiply that quotient by a^2 and see if the result is is an integer. So for Row10 that quotient would be 60^8/41990400=(1679616*10^8)/(419904*10^2)=(1679616*10^6)/(419904)=4*10^6, mentioned above; and 24611521*(4*10^6) is 98446084*10^6; a very large integer; thus, it would be 8 sexigesimal places or less. Note: a2^/b^2 has been converted into (98446084*10^6)/(60^8) which is the sum of all 8 sexigesimal places. Repeat this process for the other 14 rows and get integer outcomes. Thus, it could be determined beforehand for all 15 rows, that a^2/b^2 would be exact to 8 sexigesimal fractional places or less. And for the 15 PIRTs, if you include the Row#'s, the 60 parameters are exact!

To get the 1st sexigesimal place for Row10, divide that very large integer by 60^7, (98446084*10^6)/(279936*10^7), resulting in 35.1663... ; or**35**. For the 2nd sexigesimal place, (98446084*10^6)-(35*279936*10^7)=(98446084-97977600)*10^6=468484*10^6, which is the numerator for the 2nd fraction, and (468484*10^6)/60^6=468484/46656=10.0412... ; or **10**. And a not so easy so forth for the next 6 places. And repeat for 14 other a^2/b^2. Note: The reason for the 60^7 divisor for the 1st place is that we are really dividing numerator and denominator by 60^7 which leaves 60 in the denominator. And for the 2nd place, we use 60^6 as the divisor, which leaves 3600 in the denominator.

So, are there PIRTs that would be unsuitable on this basis? Two candidates that fail the test: a=40, b=42, c=58, 1+(a^2/b^2)=1.9070295.., for Row03; and a=85, b=132, c=157, 1+(a^2/b^2)=1.4146579... for Row14. The a^2*(60^8/b^2) for the former is 1600*(1679616*10^8/1764)=1600*(952.16327...*10^8)= 152346122448979.5918... and for the later is 7225*(1679616*10^8/17424)=7225*(96.396694...*10^8)=69646611570247.9338... . As a matter of practice, if 60^8/b^2 is an integer then a^2/b^2 will be resolved in 8 places or less.

It is easy to conclude that there is some very difficult math involved in making up this tablet.

The 15 brackets just barely fit over the 1+(a^2/b^2) values; Row04 is in by 0.0026 at the top and Row12 is in by 0.0005 at the bottom.

Fourteen of the 16 bracket boundries have repeated numerals. It can be seen that the upper for Row10 can be rewrtten as 1+(6/10)+(2/90) or 1+(36/60)+(80/3600)=1+(37/60)+(20/3600), and that in sexigesimal would be

Anybody that thinks making up this test was easy, look at 1+(a^2/b^2), Row10. Get out pencil and paper and divide 24611521 by 41990400 and get a rounded off 0.58612256611034903215973174820911... . After adding the 1, this is equivalent to making the sexigesimal calculation and getting it

To get the contribution of

But how would we know beforehand that a^2/b^2 would be eight sexigesimal places or less? One way would be to see if it could have a common denominator with 60^8. To determine that, divide b^2 into 60^8, and then multiply that quotient by a^2 and see if the result is is an integer. So for Row10 that quotient would be 60^8/41990400=(1679616*10^8)/(419904*10^2)=(1679616*10^6)/(419904)=4*10^6, mentioned above; and 24611521*(4*10^6) is 98446084*10^6; a very large integer; thus, it would be 8 sexigesimal places or less. Note: a2^/b^2 has been converted into (98446084*10^6)/(60^8) which is the sum of all 8 sexigesimal places. Repeat this process for the other 14 rows and get integer outcomes. Thus, it could be determined beforehand for all 15 rows, that a^2/b^2 would be exact to 8 sexigesimal fractional places or less. And for the 15 PIRTs, if you include the Row#'s, the 60 parameters are exact!

To get the 1st sexigesimal place for Row10, divide that very large integer by 60^7, (98446084*10^6)/(279936*10^7), resulting in 35.1663... ; or

So, are there PIRTs that would be unsuitable on this basis? Two candidates that fail the test: a=40, b=42, c=58, 1+(a^2/b^2)=1.9070295.., for Row03; and a=85, b=132, c=157, 1+(a^2/b^2)=1.4146579... for Row14. The a^2*(60^8/b^2) for the former is 1600*(1679616*10^8/1764)=1600*(952.16327...*10^8)= 152346122448979.5918... and for the later is 7225*(1679616*10^8/17424)=7225*(96.396694...*10^8)=69646611570247.9338... . As a matter of practice, if 60^8/b^2 is an integer then a^2/b^2 will be resolved in 8 places or less.

It is easy to conclude that there is some very difficult math involved in making up this tablet.

Must Lie Between |

2.02222222222222 |

1.97777777777778 |

1.93333333333333 |

1.88888888888889 |

1.84444444444444 |

1.8 |

1.75555555555556 |

1.71111111111111 |

1.66666666666667 |

1.62222222222222 |

1.57777777777778 |

1.53333333333333 |

1.48888888888889 |

1.44444444444444 |

1.4 |

1.35555555555556 |

1+(a^2/b^2) | a | c | # |

1.98340277777778 | 119 | 169 | 1 |

1.94915855208869 | 3367 | 11521 | 2 |

1.91880212673611 | 4601 | 6649 | 3 |

1.89458124005487 | 12709 | 18541 | 4 |

1.81500771604938 | 65 | 97 | 5 |

1.78519290123457 | 319 | 481 | 6 |

1.71998367626886 | 2291 | 3541 | 7 |

1.6927734375 | 799 | 1249 | 8 |

1.64266944444444 | 541 | 769 | 9 |

1.58612256611035 | 4961 | 8161 | 10 |

1.5625 | 45 | 75 | 11 |

1.48941684027778 | 1679 | 2929 | 12 |

1.45001736111111 | 25921 | 289 | 13 |

1.43023882030178 | 1771 | 3229 | 14 |

1.38716049382716 | 56 | 53 | 15 |

b_decimal | b_sexigesimal |

120 | 02;00 |

1102.02?? 3456 | 57;36 |

4800 | 01;20;00 |

13500 | 03;45;00 |

72 | 01;12 |

360 | 06;00 |

2700 | 45;00 |

960 | 16;00 |

546.52???? 600 | 10;00 |

6480 | 01;48;00 |

60 | 01;00 |

2400 | 40;00 |

?????????? 240 | 04;00 |

2700 | 45;00 |

??????????? 90 | 01;30 |

csc^2=1+(a^2/b^2) is rounded to 14 decimal places. Errors are underlined.

Columnb_sexigesimal of the student scratchpad deserves attention. The 1's column is blank except for 3 entries which add up to 01;18. Are the**1** and **8** being emphasized for a reason?

Columnb_decimal transcription of the student sexigesimal scratchpad also deserves attention. 12 of the 15 entries are multiples of 60; three that are not: 3456, 72, and 90. Are the**1** and **8** missing for a reason?

When Albert Einstein learned about "quantum entanglement", he reacted by saying it was "spooky". That exactly describes those two paragraphs above, taken together.

Columnb_sexigesimal of the student scratchpad deserves attention. The 1's column is blank except for 3 entries which add up to 01;18. Are the

Columnb_decimal transcription of the student sexigesimal scratchpad also deserves attention. 12 of the 15 entries are multiples of 60; three that are not: 3456, 72, and 90. Are the

When Albert Einstein learned about "quantum entanglement", he reacted by saying it was "spooky". That exactly describes those two paragraphs above, taken together.

Now, looking at the decimal, uncorrected version; we can get some idea of what the students faced. Get out paper and pencil and follow along with the student doing the calculations involved. I'll be using a Scientific Calculator, Sharp EL-509.

Students are to check each row to make sure the parameters are valid for a positive integer right triangle (PIRT) that fits in its bracket. If they find a mistake, then they fix it.

The toughest thing a Teacher could do, would be to toss it out and say: Go to work! But a kinder thing might be: There are six mistakes. Only one to a row. Each column has two mistakes. One obvious, one not so obvious. It's better to work from top to bottom. And mind your (b)'s!

Going from top to bottom and looking for obvious mistakes, we can see that in Row04, Colunn1, the 1+(a^2/b^2) is above its bracket. (This is in the damaged section of the tablet and this mistake is inferred by the kind and location of the other 5 mistakes; Row04 is picked for an obvious column1 mistake because the large (a) and (c) values make the fix difficult). FIND1.

Going down to Row13, we see that (a) is greater than (c), impossible, (a) or (c) is wrong. FIND2.

And at Row15 we have another (a) greater than (c), (a) or (c) is wrong. FIND3.

Now for not so obvious mistakes. At Row01 a/c=119/169=0.704 looks OK. But at Row02 a/c=0.292; now our student knows that at 45deg a/c=0.707, at 30deg

a/c=0.500 and at 15deg a/c=0.259. Then, it follows that the a/c of 0.292 (17deg) at Row02 would belong on the 30deg to 16deg tablet, if one exists, and if this is a valid PIRT. Either (a) or (c) in Row02 is wrong. FIND4.

There is one more a/c mistake, we keep ckecking the a/c's until in Row09, we find an a/c of 0.704; can't be, we already have that fraction for Row01. FIND5.

Since we have located only one mistake in Column1, that's where the last mistake must be. So, we have 10 rows of a^2/b^2 to check. Now, let's make a lucky guess (which saves a lot of work) that the Column**1** mistake we're looking for is buried in the middle at Row0**8**. FIND6.

Let's fix this mistake. We will use the equation,__a^2/b^2=a^2/(c^2-a^2)__, and plug in the (a) and (c) parameters for Row08. The solution proceeds as follows: a^2/b^2=799^2/(1249^2-799^2)=638401/(1560001-638401)=638401/921600=__0.6927094...__, the correct answer. And the tablet has for Column1 at Row08: a^2/b^2=0.6927734375, close but still a mistake! FIX1.

Let's fix the mistake in Column1 at Row04 using the same equation,__a^2/b^2=a^2/(c^2-a^2)__=12709^2/(18541^2-12709^2)=__0.8862479...__, the correct answer. And the tablet has 0.89458124..., a mistake! FIX2.

Now we have 4 mistakes in columns 2 and 3 to fix. We are going to use 1 of 2 equations to find (b): setting__a^2/b^2=x__ and substituting for a^2/b^2 in __c^2/b^2=1+(a^2/b^2)__, we have __c^2/b^2=1+x__, and we get x from Column1 in which all values are now correct, and use either __b^2=a^2/(x)__ or __b^2=c^2/(1+x)__.

For Row02 let's say (c) is wrong and go looking for (c); a=3367, a^2=11336689 and x=0.9491586... using__b^2=a^2/(x)__, we get b^2=11336689/0.9491586=11943936, b=sqr(11943936)=3456, and since __c^2=a^2+b^2__=11336689+11943936=23280625, c=sqr(23280625)=__4825__, then c=11521 is a mistake! FIX3.

For Row09 let's say (a) is wrong c=769, c^2=591361 and (1+x)=1.6426694...; using__b^2=c^2/(1+x)__, we get b^2=769^2/1.6426694=591361/1.6426694=360000, b=sqr(360000)=600, and since __a^2=c^2-b^2__=591361-360000=231361, a=sqr(231361)=__481__, then a=541 is a mistake! FIX4.

For Row13 let's say (a) is wrong; c=289, c^2=83521 and (1+x)=1.4500174... using__b^2=c^2/(1+x)__, we get b^2=289^2/1.4500174=83521/1.4500174=57600, b=sqr(57600)=240, and since __a^2=c^2-b^2__=83521-57600=25921, a=sqr(25921)=__161__, then a=25921 is a mistake! FIX5.

Now we come to the last mistake, (a) or (c) in Row15 and because we already have two (a) mistakes and only one (c) mistake, we know this is going to be a (c) mistake. (A good thing to know because a=28 and c=53 is also a valid solution for this row). So for Row15 (c) is wrong; a=56, a^2=3136 and x=0.3871604... using__b^2=a^2/(x)__, we get b^2=3136/0.3871604=8100, b=sqr(8100)=90, and since __c^2=a^2+b^2__=3136+8100=11236, and c=sqr(11236)=__106__, then c=53 is a mistake! FIX6. Done!

A PIRT sieving program will generate Table3. The program is triangletrk.html by T.L.O'Donnell. This program will generate, and sieve, positive integer right triangles, PIRT's. Quantity can be selected, as can the type of triangle to be analyzed. There are 5 types: sqeven, sqodd, regeven, regodd, and all of the above. Triangles are divided into: A>45deg, and A<45deg. These are subdivided: primitive, and similar. Primitive, means one side is a prime

number, sprm; or two sides are prime numbers, dprm; or all three sides can be divided, but not by the same number, divs. Similar, siml, means the triangle can be made into a smaller triangle of the same shape by dividing all three sides by the same number; the largest common factor, lcf, is listed. (c-b) is referred to as the track and is the index for the program; t=(c-b). When t=1, all PIRT's are primitive. When t>1, PIRT's can be primitive, lcf=1 or similar, lcf>1. The aspacing (asp) is uniform for each individual track (t). When (t) is sqeven or sqodd then the aspacing is__asp=2*sqr(t)__. A general formula for (a) is __a=t+(anum*asp)__. A general formula for (b) is __b=(a-t)*(a+t)/t/2__. And for (c) is __c=b+t__. Triangles can be selected for display and printout. The results are messaged as a summary.

__TABLE3__, PIRT Sieving Results(1+(a^2/b^2)truncated at 7 decimal places)

Selection Criteria are (c-b)=sqodd and lcf=1 and a^2/b^2<46/45 and a^2/b^2>16/45 and b/60=int(b/60) and 60^8/b^2=int(60^8/b^2)

Row01_1.9834027_divs_ c-b=49_ a=119_ b=120_ c=169_ lcf=1_ asp=14_ anum5

Row13_1.4500173_divs_ c-b=49_ a=161_ b=240_ c=289_ lcf=1_ asp=14_ anum8

Row06_1.7851929_divs_ c-b=121_ a=319_ b=360_ c=481_ lcf=1_ asp=22_ anum9

Row09_1.6426694_sprm_ c-b=169_ a=481_ b=600_ c=769_ lcf=1_ asp=26_ anum12

Row08_1.6927094_sprm_ c-b=289_ a=799_ b=960_ c=1249_ lcf=1_ asp=34_ anum15

Row12_1.4894168_divs_ c-b=529_ a=1679_ b=2400_ c=2929_ lcf=1_ asp=46_ anum25

Row14_1.4302388_sprm_ c-b=529_ a=1771_ b=2700_ c=3229_ lcf=1_ asp=46_ anum27

Row07_1.7199836_sprm_ c-b=841_ a=2291_ b=2700_ c=3541_ lcf=1_ asp=58_ anum25

Row10_1.5861225_sprm_ c-b=1681_ a=4961_ b=6480_ c=8161_ lcf=1_ asp=82_ anum40

Row03_1.9188021_divs_ c-b=1849_ a=4601_ b=4800_ c=6649_ lcf=1_ asp=86_ anum32

Row04_1.8862479_sprm_ c-b=5041_ a=12709_ b=13500_ c=18541_ lcf=1_ asp=142_ anum54

Count=11 Row# prefixes tacked on after sieving is completed.

Total tracks=6000 Right triangles/track=60 Right triangle universe=360000

Track type sqeven=38 sqodd=39 regeven=2962 regodd=2961

Track type selected=sqodd Right triangles examined=2340 aspacing=2Xsqr(c-b)

A>45deg=1056 dprime=0 sprime=325 divisible=542 similar=189

A<45deg=1284 dprime=10 sprime=402 divisible=622 similar=250

The Teacher in the class on The Theory of Numbers wanted a table in which (1+a^2/b^2) ran decrementally from 45deg to 31deg in 15 steps. But he comes up 4 short. Well, there's a saying at the school: if you get handed a lemon, start making lemonade. So, dropping lcf=1 and b/60=int(b/60) requirements; making (c-b)=sqodd or sqeven; he needs 4 PIRT's that have 60^8/b^2=int(60^8/b^2) and (c-b) equals a perfect square. He knows a PIRT that will fit in Row02, and that gives him an idea. He already knows that the Find-Fix solution will involve a lot of student effort unless they know which Row# in Column1 has the__not so obvious mistake__ to Find and Fix. Why not leave them clues in Columnb, which is on the student scratchpad. Fill in Row02 with (c-b)=1369 and b=3456. Add Row05 with (c-b)=25 and b=72. And then Row15 with (c-b)=16 and b=90, rounds out the clues, leaving **1** and **8** missing in Columnb. But what about Row11? The classic 3,4,5 PIRT fits the bracket; why not multiply it by 225 and get (c-b)=225 and b=900; making Row02, Row05, and Row15 the only ones in Columnb with b not a multiple of 60. Then he gets a really brillant idea: no, multiply it by 15 instead, so (c-b)=**15** and b=60, showing that even the Teacher can make a mistake for bright students to Find.

__TABLE3__, Continued

Row02_1.9491585_divs_ c-b=1369_ a=3367_ b=3456_ c=4825_ lcf=1_ asp=74_ anum27

Row05_1.8150077_sprm_ c-b=25_ a=65_ b=72_ c=97_ lcf=1_ asp=10_ anum4

Row11_1.5625____siml_ c-b=15_ a=45_ b=60_ c=75_ lcf=15_ asp=30_ anum1

Row15_1.3871604_siml_ c-b=16_ a=56_ b=90_ c=106_ lcf=2_ asp=8_ anum5

count=4

The Teacher that made up this student Find-Fix exercise didn't have a computer running an algorithm that would sieve out the 11 PIRT's in Table3. So, how did he do it? First he was very knowledgeable about right triangles. Second he had students he could put to work. Third he had lots of time. What knowledge did he use? He knew that with a track total of 6000, and selecting for square-odd-prime he would be examining tracks from 3sqrd up to 77sqd of which there are 20 squared odd primes. He knew from experience, if he used Sqoddprime for the track then the PIRT's would all be__primitive__ (lcf=1) with exceptions that would be eliminated by the other requirements; and also that equations __a=t+2*v*sqr(t)__ and __b=2*v^2+2*v*sqr(t)__ would apply; v=anum. These equations would allow him to determine a/b for right triangles of interest by using a/b=(t+2*v*sqr(t))/(2*v^2+2*v*sqr(t)). For 45deg, a/b=1, and solving for v: __v1=sqr(t)/sqr(2)__; and for 30deg, a/b=1/sqr(3), and __v2=(1+sqr(3))*sqr(t)/2__. Let's take a "for instance". Let's say the track is Sqoddprime, t=529=23^2, and PIRT number, v, inceases by 1 as (a) increases by asp. For each track, v=1 for largest angle A and as v increments, angle A gets smaller. Angle A is of interest between 45deg and 30deg, that is from a/b=1 down to a/b=1/sqr(3). Solving a/b=1 for 45deg, v1=sqr(t)/sqr(2)=23/1.4142=16.26; and solving a/b=1/sqr(3) for 30deg, v2=(1+sqr(3))*sqr(t)/2 =2.73*23/2=31.42. So for t=529, the PIRT's for v=16 thru v=31 are of interest, 16 in all. Which ones have b multiples of 60? Substituting sqr(t)=23 and then dividing the b formula by 60 gives us 60multiple=b/60=2*v*(v+23)/60; for v1=16, 60mult=1248/60=20.8; for v2=31, 60mult=3520/60=55.8. And in between are twelve more 60mult non-integers and two 60mult integers: for v=25, 60mult=2400/60=40 and for v=27, 60mult=2700/60=45. So for the Sqoddprime=529, we have selected two 60mult PIRT's: b=2400 and b=2700. How about their 60^8/b^2? Are they also integers? For v=25 and b=2400, 60^8/b^2=(6^8"10^8/b/b)=(6^4*10^4/24/10^2)^2=(1296*10^2/24)^2=5400^2, an integer; and for v=27 and b=2700, 60^8/b^2=4800^2, an integer. Two PIRT's meeting the Teacher's criteria have been selected. 19 of the 20 tracks are left to examine and find the other 9 PIRT's (still 4 short). So, although the work is tedious, it can be accomplished by supervised students.

The Teacher's selection method has been incorporated into an algorithm. It will generate Table4. The criteria for PIRT's are (c-b) is the square of an odd prime, angle A is between 30deg and 45 deg, b is an integer multiple of 60, and 60^8 is an integer multiple of b^2.

__TABLE4__ Teacher's Selection Method

Selection Criteria are (c-b)=sqoddprime(3sqd thru 77sqd) and anum>sqr(t)/sqr(2)and anum<(1+sqr(3))*sqr(t)/2 and b/60=int(b/60) and 60^8/b^2=int(60^8/b^2)

Row01_1.9834027_divs_ c-b=49_ a=119_ b=120_ c=169_ lcf=1_ asp=14_ anum5

Row13_1.4500173_divs_ c-b=49_ a=161_ b=240_ c=289_ lcf=1_ asp=14_ anum8

Row06_1.7851929_divs_ c-b=121_ a=319_ b=360_ c=481_ lcf=1_ asp=22_ anum9

Row09_1.6426694_sprm_ c-b=169_ a=481_ b=600_ c=769_ lcf=1_ asp=26_ anum12

Row08_1.6927094_sprm_ c-b=289_ a=799_ b=960_ c=1249_ lcf=1_ asp=34_ anum15

Row12_1.4894168_divs_ c-b=529_ a=1679_ b=2400_ c=2929_ lcf=1_ asp=46_ anum25

Row14_1.4302388_sprm_ c-b=529_ a=1771_ b=2700_ c=3229_ lcf=1_ asp=46_ anum27

Row07_1.7199836_sprm_ c-b=841_ a=2291_ b=2700_ c=3541_ lcf=1_ asp=58_ anum25

Row10_1.5861225_sprm_ c-b=1681_ a=4961_ b=6480_ c=8161_ lcf=1_ asp=82_ anum40

Row03_1.9188021_divs_ c-b=1849_ a=4601_ b=4800_ c=6649_ lcf=1_ asp=86_ anum32

Row04_1.8862479_sprm_ c-b=5041_ a=12709_ b=13500_ c=18541_ lcf=1_ asp=142_ anum54

Row08_1.6685522_divs_ c-b=5041_ a=14129_ b=17280_ c=22321_ lcf=1_ asp=142_ anum64

Count=12 Row# prefixes tacked on after sieving is completed.

Total tracks=20 Right triangles/track=23.35 Right triangle universe=467

Track type sqeven=0 sqodd=20 regeven=0 regodd=0

Track type selected=sqodd Right triangles examined=467 aspacing=2Xsqr(c-b)

A>45deg=20 dprime=0 sprime=11 divisible=9 similar=0

A<45deg=447 dprime=0 sprime=132 divisible=295 similar=20

The last PIRT repeats a bracket atready filled and will be ignored. There were 467 PIRT's to examine, still tedious, but doable.

Let's do 31 more tracks using the Teacher's sieving method. Which will take us out to where c>60^3. And since space on the tablet limits us to 3 sexigesimal integer places; this is a good place to stop.

Teacher's Selection Method Continued

Selction Criteria are (c-b)=sqoddprime(79sqd thru 239sqd) and anum>sqr(t)/sqr(2)and anum<(1+sqr(3))*sqr(t)/2 and b/60=int(b/60) and 60^8/b^2=int(60^8/b^2)

Row11_1.5395333_divs_ c-b=6241_ a=19039_ b=25920_ c=32161_ lcf=1_ asp=158

_ anum81

Row16_1.3533848_divs_ c-b=9409_ a=34241_ b=57600_ c=67009_ lcf=1_ asp=194

_ anum128

Row04_1.8741990_divs_ c-b=26569_ a=67319_ b=72000_ c=98569_ lcf=1_ asp=326

_ anum125

Count=3 Row# prefixes tacked on after sieving is completed.

Total tracks=31 Right triangles/track=103.71 Right triangle universe=3215

Track type sqeven=0 sqodd=31 regeven=0 regodd=0

Track type selected=sqodd Right triangles examined=3215 aspacing=2Xsqr(c-b)

A>45deg=31 dprime=0 sprime=10 divisible=21 similar=0

A<45deg=3215 dprime=0 sprime=715 divisible=2438 similar=31

The 1st PIRT for Row11 would have been the 12th selected. The 2nd is below this Table; the 3rd repeats a bracket already filled; 2nd and 3rd will be ignored. The summary shows that there were 3215 PIRT's left to examine for selection, if the Teacher had continued the sieving. But by doing only one more track he would have found Row11!

__Internet Links__

http://en.wikipedia.org/wiki/Plimpton_322

http://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Robson105-120.pdf by Eleanor Robson

http://www.math.ubc.ca/~cass/courses/m446-03/pl322/pl322.html by Bill Casselman

http://aleph0.clarku.edu/~djoyce/mathhist/plimpnote.html by David E. Joyce

http://www.math.stonybrook.edu/~tony/archive/336S06/plimpton322.html by Anthony Phillips

http://arxiv.org/pdf/1004.0025v1.pdf by Abdulrahman A. Abdulaziz

http://arxiv.org/pdf/1109.3814v2.pdf by Anthony Phillips

email comments to T.L.O'Donnell, plimpton322@localnet.com

Students are to check each row to make sure the parameters are valid for a positive integer right triangle (PIRT) that fits in its bracket. If they find a mistake, then they fix it.

The toughest thing a Teacher could do, would be to toss it out and say: Go to work! But a kinder thing might be: There are six mistakes. Only one to a row. Each column has two mistakes. One obvious, one not so obvious. It's better to work from top to bottom. And mind your (b)'s!

Going from top to bottom and looking for obvious mistakes, we can see that in Row04, Colunn1, the 1+(a^2/b^2) is above its bracket. (This is in the damaged section of the tablet and this mistake is inferred by the kind and location of the other 5 mistakes; Row04 is picked for an obvious column1 mistake because the large (a) and (c) values make the fix difficult). FIND1.

Going down to Row13, we see that (a) is greater than (c), impossible, (a) or (c) is wrong. FIND2.

And at Row15 we have another (a) greater than (c), (a) or (c) is wrong. FIND3.

Now for not so obvious mistakes. At Row01 a/c=119/169=0.704 looks OK. But at Row02 a/c=0.292; now our student knows that at 45deg a/c=0.707, at 30deg

a/c=0.500 and at 15deg a/c=0.259. Then, it follows that the a/c of 0.292 (17deg) at Row02 would belong on the 30deg to 16deg tablet, if one exists, and if this is a valid PIRT. Either (a) or (c) in Row02 is wrong. FIND4.

There is one more a/c mistake, we keep ckecking the a/c's until in Row09, we find an a/c of 0.704; can't be, we already have that fraction for Row01. FIND5.

Since we have located only one mistake in Column1, that's where the last mistake must be. So, we have 10 rows of a^2/b^2 to check. Now, let's make a lucky guess (which saves a lot of work) that the Column

Let's fix this mistake. We will use the equation,

Let's fix the mistake in Column1 at Row04 using the same equation,

Now we have 4 mistakes in columns 2 and 3 to fix. We are going to use 1 of 2 equations to find (b): setting

For Row02 let's say (c) is wrong and go looking for (c); a=3367, a^2=11336689 and x=0.9491586... using

For Row09 let's say (a) is wrong c=769, c^2=591361 and (1+x)=1.6426694...; using

For Row13 let's say (a) is wrong; c=289, c^2=83521 and (1+x)=1.4500174... using

Now we come to the last mistake, (a) or (c) in Row15 and because we already have two (a) mistakes and only one (c) mistake, we know this is going to be a (c) mistake. (A good thing to know because a=28 and c=53 is also a valid solution for this row). So for Row15 (c) is wrong; a=56, a^2=3136 and x=0.3871604... using

A PIRT sieving program will generate Table3. The program is triangletrk.html by T.L.O'Donnell. This program will generate, and sieve, positive integer right triangles, PIRT's. Quantity can be selected, as can the type of triangle to be analyzed. There are 5 types: sqeven, sqodd, regeven, regodd, and all of the above. Triangles are divided into: A>45deg, and A<45deg. These are subdivided: primitive, and similar. Primitive, means one side is a prime

number, sprm; or two sides are prime numbers, dprm; or all three sides can be divided, but not by the same number, divs. Similar, siml, means the triangle can be made into a smaller triangle of the same shape by dividing all three sides by the same number; the largest common factor, lcf, is listed. (c-b) is referred to as the track and is the index for the program; t=(c-b). When t=1, all PIRT's are primitive. When t>1, PIRT's can be primitive, lcf=1 or similar, lcf>1. The aspacing (asp) is uniform for each individual track (t). When (t) is sqeven or sqodd then the aspacing is

Selection Criteria are (c-b)=sqodd and lcf=1 and a^2/b^2<46/45 and a^2/b^2>16/45 and b/60=int(b/60) and 60^8/b^2=int(60^8/b^2)

Row01_1.9834027_divs_ c-b=49_ a=119_ b=120_ c=169_ lcf=1_ asp=14_ anum5

Row13_1.4500173_divs_ c-b=49_ a=161_ b=240_ c=289_ lcf=1_ asp=14_ anum8

Row06_1.7851929_divs_ c-b=121_ a=319_ b=360_ c=481_ lcf=1_ asp=22_ anum9

Row09_1.6426694_sprm_ c-b=169_ a=481_ b=600_ c=769_ lcf=1_ asp=26_ anum12

Row08_1.6927094_sprm_ c-b=289_ a=799_ b=960_ c=1249_ lcf=1_ asp=34_ anum15

Row12_1.4894168_divs_ c-b=529_ a=1679_ b=2400_ c=2929_ lcf=1_ asp=46_ anum25

Row14_1.4302388_sprm_ c-b=529_ a=1771_ b=2700_ c=3229_ lcf=1_ asp=46_ anum27

Row07_1.7199836_sprm_ c-b=841_ a=2291_ b=2700_ c=3541_ lcf=1_ asp=58_ anum25

Row10_1.5861225_sprm_ c-b=1681_ a=4961_ b=6480_ c=8161_ lcf=1_ asp=82_ anum40

Row03_1.9188021_divs_ c-b=1849_ a=4601_ b=4800_ c=6649_ lcf=1_ asp=86_ anum32

Row04_1.8862479_sprm_ c-b=5041_ a=12709_ b=13500_ c=18541_ lcf=1_ asp=142_ anum54

Count=11 Row# prefixes tacked on after sieving is completed.

Total tracks=6000 Right triangles/track=60 Right triangle universe=360000

Track type sqeven=38 sqodd=39 regeven=2962 regodd=2961

Track type selected=sqodd Right triangles examined=2340 aspacing=2Xsqr(c-b)

A>45deg=1056 dprime=0 sprime=325 divisible=542 similar=189

A<45deg=1284 dprime=10 sprime=402 divisible=622 similar=250

The Teacher in the class on The Theory of Numbers wanted a table in which (1+a^2/b^2) ran decrementally from 45deg to 31deg in 15 steps. But he comes up 4 short. Well, there's a saying at the school: if you get handed a lemon, start making lemonade. So, dropping lcf=1 and b/60=int(b/60) requirements; making (c-b)=sqodd or sqeven; he needs 4 PIRT's that have 60^8/b^2=int(60^8/b^2) and (c-b) equals a perfect square. He knows a PIRT that will fit in Row02, and that gives him an idea. He already knows that the Find-Fix solution will involve a lot of student effort unless they know which Row# in Column1 has the

Row02_1.9491585_divs_ c-b=1369_ a=3367_ b=3456_ c=4825_ lcf=1_ asp=74_ anum27

Row05_1.8150077_sprm_ c-b=25_ a=65_ b=72_ c=97_ lcf=1_ asp=10_ anum4

Row11_1.5625____siml_ c-b=15_ a=45_ b=60_ c=75_ lcf=15_ asp=30_ anum1

Row15_1.3871604_siml_ c-b=16_ a=56_ b=90_ c=106_ lcf=2_ asp=8_ anum5

count=4

The Teacher that made up this student Find-Fix exercise didn't have a computer running an algorithm that would sieve out the 11 PIRT's in Table3. So, how did he do it? First he was very knowledgeable about right triangles. Second he had students he could put to work. Third he had lots of time. What knowledge did he use? He knew that with a track total of 6000, and selecting for square-odd-prime he would be examining tracks from 3sqrd up to 77sqd of which there are 20 squared odd primes. He knew from experience, if he used Sqoddprime for the track then the PIRT's would all be

The Teacher's selection method has been incorporated into an algorithm. It will generate Table4. The criteria for PIRT's are (c-b) is the square of an odd prime, angle A is between 30deg and 45 deg, b is an integer multiple of 60, and 60^8 is an integer multiple of b^2.

Selection Criteria are (c-b)=sqoddprime(3sqd thru 77sqd) and anum>sqr(t)/sqr(2)and anum<(1+sqr(3))*sqr(t)/2 and b/60=int(b/60) and 60^8/b^2=int(60^8/b^2)

Row01_1.9834027_divs_ c-b=49_ a=119_ b=120_ c=169_ lcf=1_ asp=14_ anum5

Row13_1.4500173_divs_ c-b=49_ a=161_ b=240_ c=289_ lcf=1_ asp=14_ anum8

Row06_1.7851929_divs_ c-b=121_ a=319_ b=360_ c=481_ lcf=1_ asp=22_ anum9

Row09_1.6426694_sprm_ c-b=169_ a=481_ b=600_ c=769_ lcf=1_ asp=26_ anum12

Row08_1.6927094_sprm_ c-b=289_ a=799_ b=960_ c=1249_ lcf=1_ asp=34_ anum15

Row12_1.4894168_divs_ c-b=529_ a=1679_ b=2400_ c=2929_ lcf=1_ asp=46_ anum25

Row14_1.4302388_sprm_ c-b=529_ a=1771_ b=2700_ c=3229_ lcf=1_ asp=46_ anum27

Row07_1.7199836_sprm_ c-b=841_ a=2291_ b=2700_ c=3541_ lcf=1_ asp=58_ anum25

Row10_1.5861225_sprm_ c-b=1681_ a=4961_ b=6480_ c=8161_ lcf=1_ asp=82_ anum40

Row03_1.9188021_divs_ c-b=1849_ a=4601_ b=4800_ c=6649_ lcf=1_ asp=86_ anum32

Row04_1.8862479_sprm_ c-b=5041_ a=12709_ b=13500_ c=18541_ lcf=1_ asp=142_ anum54

Row08_1.6685522_divs_ c-b=5041_ a=14129_ b=17280_ c=22321_ lcf=1_ asp=142_ anum64

Count=12 Row# prefixes tacked on after sieving is completed.

Total tracks=20 Right triangles/track=23.35 Right triangle universe=467

Track type sqeven=0 sqodd=20 regeven=0 regodd=0

Track type selected=sqodd Right triangles examined=467 aspacing=2Xsqr(c-b)

A>45deg=20 dprime=0 sprime=11 divisible=9 similar=0

A<45deg=447 dprime=0 sprime=132 divisible=295 similar=20

The last PIRT repeats a bracket atready filled and will be ignored. There were 467 PIRT's to examine, still tedious, but doable.

Let's do 31 more tracks using the Teacher's sieving method. Which will take us out to where c>60^3. And since space on the tablet limits us to 3 sexigesimal integer places; this is a good place to stop.

Teacher's Selection Method Continued

Selction Criteria are (c-b)=sqoddprime(79sqd thru 239sqd) and anum>sqr(t)/sqr(2)and anum<(1+sqr(3))*sqr(t)/2 and b/60=int(b/60) and 60^8/b^2=int(60^8/b^2)

Row11_1.5395333_divs_ c-b=6241_ a=19039_ b=25920_ c=32161_ lcf=1_ asp=158

_ anum81

Row16_1.3533848_divs_ c-b=9409_ a=34241_ b=57600_ c=67009_ lcf=1_ asp=194

_ anum128

Row04_1.8741990_divs_ c-b=26569_ a=67319_ b=72000_ c=98569_ lcf=1_ asp=326

_ anum125

Count=3 Row# prefixes tacked on after sieving is completed.

Total tracks=31 Right triangles/track=103.71 Right triangle universe=3215

Track type sqeven=0 sqodd=31 regeven=0 regodd=0

Track type selected=sqodd Right triangles examined=3215 aspacing=2Xsqr(c-b)

A>45deg=31 dprime=0 sprime=10 divisible=21 similar=0

A<45deg=3215 dprime=0 sprime=715 divisible=2438 similar=31

The 1st PIRT for Row11 would have been the 12th selected. The 2nd is below this Table; the 3rd repeats a bracket already filled; 2nd and 3rd will be ignored. The summary shows that there were 3215 PIRT's left to examine for selection, if the Teacher had continued the sieving. But by doing only one more track he would have found Row11!

http://en.wikipedia.org/wiki/Plimpton_322

http://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Robson105-120.pdf by Eleanor Robson

http://www.math.ubc.ca/~cass/courses/m446-03/pl322/pl322.html by Bill Casselman

http://aleph0.clarku.edu/~djoyce/mathhist/plimpnote.html by David E. Joyce

http://www.math.stonybrook.edu/~tony/archive/336S06/plimpton322.html by Anthony Phillips

http://arxiv.org/pdf/1004.0025v1.pdf by Abdulrahman A. Abdulaziz

http://arxiv.org/pdf/1109.3814v2.pdf by Anthony Phillips

email comments to T.L.O'Donnell, plimpton322@localnet.com