Show that the area A of a triangle whose sides form a pythagorean triple is always divisible by 6.

Hint: For an integer x, x^2 is of the form 4k or 1+8k, and when 3 doesn't divide x, x^2 is of the form 1+3k.

I don't know what to do. I figured the strategy would be to show that it is divisible by 2 and also by 3, so that it is divisible by 6=2*3, but I don't know how to go about this or if it is a correct approach.

thanks