1. ## pythagorean triangles.

prove 12 divides product of legs

prove 60 divides product of sides

2. Originally Posted by stumped765
prove 12 divides product of legs

prove 60 divides product of sides

What are you talking about, anyway? And are you sure this is university's stuff, number theory's?

Tonio

3. ## exercise in mod arithmetic

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?

4. You are quite right !

5. im stuck on the mod 5 when u=1(5) and v=2(5) or v=3(5)

6. Originally Posted by qmech
$\displaystyle (u^2-v^2)(2uv)(u^2+v^2)$
Originally Posted by stumped765
im stuck on the mod 5 when u=1(5) and v=2(5) or v=3(5)
if u = 1(5) and v = 2(5) then u^2 + v^2 = 1^2 + 2^2 = 1 + 4 = 5 = 0(5).
if u = 1(5) and v = 3(5) then u^2 + v^2 = 1^2 + 3^2 = 1 + 9 = 10 = 0(5).