### Babylonian Pythagorean Triplets

*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).* an even power of
2, the second an even power of 3.* (The order in the difference term could of course be reversed, depending
on which one's bigger).

In fact, notice that 5 : 12 : 13 and 7 : 24 : 25 are already of this form, with 2^{2},
3^{2} and 2^{4}, 3^{2}. What about 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.