A classic mystery locked in a 3,600-year-old Babylonian clay tablet has been solved! How did the Babylonians know the Pythagorean theorem a thousand years before the Greek mathematician and philosopher was born? For those who have forgotten their geometry, the Pythagorean theorem states: "The square of the hypotenuse of a right-angle triangle is equal to the sum of the squares of the two sides."
In October, 1978, I was on an American Airlines flight to Salt Lake City. The stewardess handed me Scientific American. Page 109 of an article contained something that would change my life. It was a Babylonian Clay Cuneiform Tablet dated about 1600 B.C. It is one of the oldest-known documents concerned with number theory. In disguised form, the tablet contains 15 sets of Pythagorean Triples; they are positive, whole numbers. The tenth set yields 4,961, 6,480, 8,161. This means 4,961 x 4,961 + 6,480 x 6,480 = 8,161 x 8,161,
The relationship, a2 + b2 = c2 for whole numbers like (3,4,5), is called a Pythagorean Triple. Even with the formula, it has been very difficult to calculate exact triples. Note that a triple (a,b,c), of natural numbers is called a Pythagorean triple if a2 + b2 = c2. (3,4,5) is a Pythagorean triple, but (1,1,2) and (1,2,3) are not.
The clay tablet is called Plimpton #322 after its location in the Plimpton Collection of Columbia University. It is in disguised form.
For decades, mathematicians have been trying to determine how this could be. The solution depended on history, geography, and art &emdash; as well as mathematics.
I reasoned that the City of Babylon, capital of Mesopotamia, was laid out as a rectangle, and rectangles can be turned into two triangles by drawing a diagonal line from one corner to another. Squares are equal-sided rectangles, and for the Babylonians who were concerned with area measurements of city blocks and properties, the formula was b2 = c2 -a2 = (c-a) . (c+a), exactly corresponding to the theorem later established by Pythagoras.
From this reasoning, I deduced that the Babylonians were able to produce their perfect triples as 16=25-9 by rearranging a basic four-by-four square.
Could a triangle be converted to a rectangle? The simplest right triangle is (3,4,5). I added a Phantom Square; it's imaginary! Notice that 5 and 3 meet at a point and if I rotate the hypotenuse around this point, it suggests small squares below the (t-s) (t+s) rectangle. Adding in the four small squares that aren't there gives rise to the 2 times 2 Phantom Square.
If we had rotated s=3 around the same point as t meeting s, a rectangle measuring only 6 units high would be created. It is perfectly balanced and symmetrical; it is 2 units wide. On top of it is a real square measuring 2 by 2; below it is a Phantom Square measuring 2 by 2. The real and imagined areas form a rectangle for t. All of these came from r = 4 arm of the original triangle.
Using the Phantom Square, we will now show that all Pythagorean triples can be built from multiples of the (3,4,5) triangle. Consider multiplying the 4 by 4 square by 2. If we apply the Babylonian rectangular rearrangement, we start out with a square: r = 8 = 64 square units. Let's divide it by 2 as we originally did with (3.4.5). We now produce a tall, thin rectangle. It is 32 by 2. To add symmetry, the needed "Phantom Square" is only 2 by 2. However, we see 2s = 30, and 2t = 34. We have produced a new triple from the arm of the triangle measuring 8 across. Since 2s = 30, s=15, and since 2t=34, t is 17 units. The produced triple is (15,8,17) and it is a Pythagorean triple.
We repeat the approach starting with the r=8 square, but this time we divide by 4. We produce a rectangle 4 by 16. We again need a "Phantom Square" for the required balance; it is 4 by 4. Here we see 2s=12, and 2t=20; therefore s=6 and t=10. This approach generates the expected (6,8,10).
We know that this is a Pythagorean triple since it is doubling the values of 3 and 4 and 5. Consider the next attempt. Let's multiply the original 4 by 4 from (3,4,5) by the number 3. This produces a square that measures 12 by 12. We now have r=12, therefore its square is r2 = 144. This is divisible by 2,4,6 and 8 since even numbers are needed for the symmetry. Each rectangle must first have an area of 144, thereby giving heights of 72, 36, 24 and 18. Using the appropriate Phantom Square for each, yields: 2s=70, 2t=74; 2s=32, 2t=40; 2s=18, 2t=30; and 2s=10, 2t=26. We see that these resulted from the 12 time 12 square. Four Pythagorean triples are generated; they are (35,12,37), (16,12,20), (9,12,15) and (5,12,13). Many books state that triples such as (15,8,17), (35,12,37) and (5,12,13) are primitive Pythagorean triples! As such, they cannot be generated from any other triangles! However, we have just seen that by using the Babylonian approach, these "primitive" triples all came from (3,4,5)!
Given a triangle whose side measures 100 units, produce a few Pythagorean triples.
There are seven Pythagorean triples all of which contain the number 100 which is located at the center of the Pythagorean triple.
Some mathematicians will claim that four of the seven are primitive. Are they?
Yes, the Babylonians knew the concepts of the Pythagorean theorem about 1,000 years before Pythagoras lived. They used 60 as their number base. It led to 60 seconds, 60 minutes, etc. Note also that 3 x 4 x 5 equals 60. Frenicle, in France in 1676 proved that any multiplied Pythagorean triple must be exactly divisible by 60!
This article originally appeared in the 08/01/1998 issue of THE Journal.