# I can't understand this problem

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1. Aug 15, 2016

### nmego12345

1. The problem statement, all variables and given/known data
Ok so this isn't really a problem, more like a problem set, I'm not sure if I'm able to understand it yet.
The context is determining all the primitive pythogrean triples
1. Letting x = a/c and y = b/c, we see that (x, y) is a point on the unit circle with rational number coordinates. In the section on solving systems of simultaneous equations, one obtained a parametrization of all solutions of this equation. Verify that if one allows the parameter t to take on only rational solutions, then one obtains a parameterization of all the solutions which are rational numbers.
2. Determine conditions on t so that one has a parametrization of all positive rational solutions.
3. Let t = u/v where u and v are relatively prime natural numbers and v > u. Then substitute these into the parametrization to obtain

Now, it is tempting, but not valid to conclude that the numerators and denominators on each side are equal. However, one can conclude that there is a positive rational number r such that

Explain why this is the case.

2. Relevant equations

x^2 + y^2 = 1
If x,y and z are vertices of a triangle and the triangle is right at z then
(xz)^2 + (zy)^2 = (xy)^2

3. The attempt at a solution
This is my attempt at understanding and solving this problem set, correct me if i'm wrong

1. I guess this asks me to get the paramaterization for:
(a/c)^2 + (b/c)^2 = 1
let's insert in
x = a/c, y = b/c

x^2 + y^2 = 1
then it's pretty straightforward
cos(t) = x, sin(t) = y

now I have to verify that if one allows the parameter t to take on only rational solutions, then one obtains a parameterization of all the solutions which are rational numbers.

What does the problem mean by "one obtains a a parameterization of all the rational solutions"?

I think it tells me to find the t values for which x, y are rational. They are
0 + mpi/2, pi/6 + mpi/2, pi/4 + mpi/2, pi/3 + m/pi2, pi/2 + m/pi2,

where m belongs to the set ℕ U {0}

2. This is easy to understand
0 < t < pi/2

Done

3. t = u/v

0 < u/v < pi/2

Now I can't understand how do we get

That's one thing.

Another thing is when the problem says "not valid to conclude that the numerators and denominators on each side are equal"

does it means "not valid to conclude that v^2 - u^2 = 2uv"?

Thanks

2. Aug 15, 2016

### RUber

This is taking a different approach from the one you used.
Look at
$\left( \frac{a}{c}\right)^2+ \left( \frac{b}{c}\right)^2 = 1$
1 is a perfect square, so the left side must be able to be written as a square.

3. Aug 15, 2016

### nmego12345

I'm sorry, I can't understand what are you trying to explain

4. Aug 15, 2016

### RUber

The problem states:
Did the parameterization referred to use sine and cosine? I doubt that is the one the problem is asking for, since having a rational t does not guarantee that $\sin(t)$ and $\cos(t)$ are rational.

5. Aug 15, 2016

### SammyS

Staff Emeritus
What is the parametrization you are to use here?

It looks like it's
$\displaystyle x = \frac{1-t^2}{1+t^2}$

$\displaystyle y = \frac{2t}{1+t^2}$

Oh, I see that RUber beat me to it !