prove 12 divides product of legs
prove 60 divides product of sides
The general legs of an integral sided right triangle are:
$\displaystyle u^2-v^2,2uv,u^2+v^2$
Let's look at the product of the legs:
$\displaystyle (u^2-v^2)( 2uv)=2uv(u+v)(u-v)$
We want to see if we can find the factors 2,2 and 3 in here. If so, the product is divisible by 12.
We have 1 factor of 2 as a coefficient, so that was easy.
Where is the 2nd factor of 2?
If either u or v is even, we have the other factor of 2.
If not, they are both odd. But then u+v is even.
So we're guaranteed to have 2 factors of 2.
What about the factor of 3?
Set up a table. Either u=0(3),u=1(3), or u=2(3), and the same for v.
If u=0(3), we're done.
If u=1(3), and v=0(3) we're done.
If u=1(3), and v=1(3) , u-v = 0(3), so we've found it.
If u=1(3), and v=2(3) , u+v = 0(3), so we've found it.
You'll see if you check the u=2(3) case that the 3 is always there too.
That solves the first problem.
The second part asks you to look at the product
$\displaystyle (u^2-v^2)(2uv)(u^2+v^2)$
and show that it has factors of 2,2,3 and 5.
Finding the factor of 5 will need you to look at all possible congruence values of u and v mod 5.
Can you continue?