If x^2 + y^2 = z^2, prove xyz is a multiple of 60

I.e. one side of every such triplet is divisible by 3, another by 4, and another by 5.

what's a side? if you're relating it to triangle, then 5,12,13 is a counter-example to what you're saying...

Note that 60=(2^2)(3)(5)=(3)(4)(5) and use modular arithmetic.
As an example let's show xyz is divisible by 4 by case analysis.
1) Suppose z=0 (mod 4). Then xyz=0 (4) trivially.
2) Suppose z=1 (4). Then z^2=1 (4) so x^2+y^2=1 (4). Hence one of x or y must be 0 (4) while the other must be 1 (4). It follows xyz=0 (4).
3) Suppose z=2 (4). Then z^2=4 (4) so x^2+y^2=0 (4). Hence x=y=0 (4) and so xyz=0 (4).
4) Suppose z=3 (4). Then z^2=9=1 (4). See 2).

mod 4???

Note that 60=(2^2)(3)(5)=(3)(4)(5) and use modular arithmetic.
As an example let's show xyz is divisible by 4 by case analysis.
1) Suppose z=0 (mod 4). Then xyz=0 (4) trivially.
2) Suppose z=1 (4). Then z^2=1 (4) so x^2+y^2=1 (4). Hence one of x or y must be 0 (4) while the other must be 1 (4). It follows xyz=0 (4).
3) Suppose z=2 (4). Then z^2=4 (4) so x^2+y^2=0 (4). Hence x=y=0 (4) and so xyz=0 (4).
4) Suppose z=3 (4). Then z^2=9=1 (4). See 2).

Mod = modulo. It's a system of counting where the numbers repeat themselves every certain interval. For example, clocks count mod 12 - they don't count the exact number of hours that have gone past since the first time it was ever 12 o'clock, they count it since the last time it was 12 o'clock. So any number n = n ± km (mod m). For example, 0 = 4 = 8 = 12 = 444 = -16 (mod 4). (It often helps to visualise it in this 'clock' way, but starting at 0, not 12. So counting up from 0, mod 12, would go: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, ...)

This type of arithmetic often helps to test for divisibility. Since the natural (counting numbers), mod 4, can easily be written:
0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2...
(0 1 2 3 4 5 6 7...)
then it's obvious that any number n = 0 (mod 4) is divisible by 4. For example, 4 = 0 (mod 4), and 4 is divisible by 4. 27 = 1 (mod 4), so is not divisible by 4 (and will, of course, leave a remainder of 1). What Gaz said is that anything divisible by 60 will be divisible by 3, 4 and 5, and hence xyz = 0 (mod 3), xyz = 0 (mod 4), and xyz = 0 (mod 5). He's proven xyz = 0 (mod 4) [for] for you - now try it for mod 3 and 5.

http://en.wikipedia.org/wiki/Modular_arithmetic probably gives a better explanation than me.

I've never really understood modulo. Can someone explain why it's -16?

Rather than (what I would consider to be) the usual, for instance, (where the (often positive) number after the congruent sign is typically chosen to be as close to zero as possible), I believe generalebriety was simply showing that any multiple of four is congruent to any other multiple of four (mod 4).

To write what he wrote in what I would consider to be the usual way, "4 = 8 = 12 = 444 = -16 = 0 (mod 4)" (but with congruent signs).

Hopefully that helps. (although I fear I haven't explained it particularly well)