prove 12 divides product of legs
prove 60 divides product of sides
The general legs of an integral sided right triangle are:
Let's look at the product of the legs:
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
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?