Lecture on Counting in Babylon

## Manufacturing Babylonian Pythagorean Triplets (a.k.a. triples)

*Michael Fowler*

*Question*: the Babylonians catalogued many Pythagorean
triplets of numbers (centuries before Pythagoras!) including the enormous 3,367
: 3,456 : 4,825. Obviously, they didn't
check every triplet of integers, even plausible looking ones, up to that value.
How could they possibly have come up with that set?

Suppose they *did* discover a few sets of integers by
trial and error, say 3 : 4: 5, 5 : 12 :
13, 7 : 24 : 25, 8 : 15 : 17. We'll assume they didn't count 6 : 8 : 10, and other triplets where all three numbers
have a common factor, since that's not really anything new.

Now they contemplate their collection of triplets. Remember,
they're focused on *sums of squares* here. So, they probably noticed that all their triplets had a
remarkable common property: the largest member of each triplet (whose *square* is of course the sum of the squares
of the other two members) is in fact *itself* a sum of two squares! Check it out: 5 = 2^{2} + 1^{2},
13 = 3^{2} + 2^{2}, 25 = 4^{2} + 3^{2}, 17 = 4^{2}
+ 1^{2}.

Staring at the triplets a little longer they might have seen
that once you express the largest member as a *sum* of two squares, one of
the other two members of the triplet is the *difference* of the same two
squares! That is, 3 = 2^{2} - 1^{2}, 5 = 3^{2}
-2^{2}, 7 = 4^{2} - 3^{2}, 15 = 4^{2} - 1^{2}.

How does the third member of the triplet relate to the numbers we squared and added to get the largest member? It's just twice their product! That is, 4 = 2x2x1, 12 = 2x3x2, 24 = 2x4x3, 8 = 2x4x1.

This at least suggests a way to manufacture larger triplets, which can then be checked by multiplication. We need to take the squares of two numbers that don't have a common factor (otherwise, all members of the triplet will have that factor).

Let's take the biggest triplet member to be the sum of an even power of 2 and an even power of 3 (so, a sum of two squares).

Another member of the triplet is then the difference of the squares, taken of course positive, and the third is twice the product (of the numbers, not the squares).

In fact, notice that 5 : 12 : 13 and 7 : 24 : 25 are already of this form, with largest members 2^{2} + 3^{2} and 2^{4} + 3^{2}. What about a triplet with largest member 2^{6} + 3^{2}? That gives 48 : 55 : 73. Try another: 2^{6} + 3^{4} gives 17 : 144 : 145. But why stop there? Let's try something bigger: 2^{12} + 3^{6}. *That* gives the triplet 3,367 : 3,456 : 4,825. Not so mysterious after all.

Finally, why are they doing this? Just for fun? Actually collections of these right-angled triangles are really early trig tables. And, with a ruler, they can reproduce angles precisely. Maybe for building? Measuring land?