# Thread: Primitive Pyth. triples: Solutions of x^2 + y^2 = 2 z^2

1. ## Primitive Pyth. triples: Solutions of x^2 + y^2 = 2 z^2

Theorem: The positive primitive solutions of $\displaystyle x^2 + y^2 = z^2$ with y even are $\displaystyle x = r^2 - s^2, y = 2rs, z = r^2 + s^2$, where r and s are arbitrary integers of opposite parity with r>s>0 and gcd(r,s)=1.

Using this theorem, find all solutions of the equation $\displaystyle x^2 + y^2 = 2z^2$
(hint: write the equation in the form $\displaystyle (x+y)^2 + (x-y)^2 = (2z)^2$)
==================================

I am not sure how to do this problem. I don't even understand how to get the equation in the hint.

I hope someone can help me out. Thank you!

[also under discussion in math links forum]

2. Originally Posted by kingwinner
Theorem: The positive primitive solutions of $\displaystyle x^2 + y^2 = z^2$ with y even are $\displaystyle x = r^2 - s^2, y = 2rs, z = r^2 + s^2$, where r and s are arbitrary integers of opposite parity with r>s>0 and (r,s)=1.

Using this theorem, find all solutions of the equation $\displaystyle x^2 + y^2 = 2z^2$
(hint: write the equation in the form $\displaystyle (x+y)^2 + (x-y)^2 = (2z)^2$)
==================================

I am not sure how to do this problem. I don't even understand how to get the equation in the hint.

I hope someone can help me out. Thank you!

[also under discussion in math links forum]
Here's help with the hint, let's see if you can take it from here.

Given $\displaystyle x^2+y^2=z^2$, lets find $\displaystyle a,b$ such that $\displaystyle (a+b)^2+(a-b)^2=(2z)^2$

First let's solve $\displaystyle (a'+b')^2+(a'-b')^2=z^2$

This means $\displaystyle \begin{cases} a'+b'=x\\a'-b'=y \end{cases} \implies \begin{cases} a'=\frac{x-y}{2}\\b'=\frac{x+y}{2} \end{cases}$.

But $\displaystyle x+y,x-y$ are both odd so let $\displaystyle a=2a'$ and $\displaystyle b=2b'$.

So we have $\displaystyle (a'+b')^2+(a'-b')^2=z^2 \implies \tfrac14(a+b)^2+\tfrac14(a-b)^2=z^2$,

or in other words $\displaystyle (a+b)^2+(a-b)^2=(2z)^2$ where $\displaystyle a=x-y$ and $\displaystyle b=x+y$.

3. But I don't think these 2 equations are exactly the same.
$\displaystyle x^2 + y^2 = 2z^2$
$\displaystyle (x+y)^2 + (x-y)^2 = (2z)^2$

Expanding the second one, I get $\displaystyle 2x^2 + 2y^2 = 4z^2$??

4. Originally Posted by kingwinner
But I don't think these 2 equations are exactly the same.
$\displaystyle x^2 + y^2 = 2z^2$
$\displaystyle (x+y)^2 + (x-y)^2 = (2z)^2$

Expanding the second one, I get $\displaystyle 2x^2 + 2y^2 = 4z^2$??
Correct, now divide both sides by $\displaystyle 2$, and what do we get?

5. Originally Posted by chiph588@
Correct, now divide both sides by $\displaystyle 2$, and what do we get?
I was looking at the wrong equation before...silly me...

Any hints about how to do the actual problem? I have no idea how to prove things like this...where should I begin?

Thanks!

6. Originally Posted by kingwinner
I was looking at the wrong equation before...silly me...

Any hints about how to do the actual problem? I have no idea how to prove things like this...where should I begin?

Thanks!
Well, we now know that $\displaystyle x^2+y^2=2z^2\iff(x+y)^2+(x-y)^2=(2z)^2$.

The last equation is of the form $\displaystyle a^2+b^2=c^2$ and we know how to solve these, right?

7. Originally Posted by chiph588@
Well, we now know that $\displaystyle x^2+y^2=2z^2\iff(x+y)^2+(x-y)^2=(2z)^2$.

The last equation is of the form $\displaystyle a^2+b^2=c^2$ and we know how to solve these, right?
Yes, but I think we're supposed to find all solutions x,y,z
In the form:
x=...
y=...
z=...

Not something like x+y=..., x-y=...?

8. Originally Posted by kingwinner
Yes, but I think we're supposed to find all solutions x,y,z
In the form:
x=...
y=...
z=...

Not something like (x+y)^2=..., (x-y)^2=...?
Right, let's use the formulas you had up above...

$\displaystyle \begin{cases} x+y=m^2-n^2\\x-y=2mn\\2z=m^2+n^2 \end{cases}$

Now solve for $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$.

One thing to note though: we need $\displaystyle m$ and $\displaystyle n$ to have the same parity. Can you see why?

9. Originally Posted by chiph588@
Right, let's use the formulas you had up above...

$\displaystyle \begin{cases} x+y=m^2-n^2\\x-y=2mn\\2z=m^2+n^2 \end{cases}$

Now solve for $\displaystyle x$, $\displaystyle y$, and $\displaystyle z$.

One thing to note though: we need $\displaystyle m$ and $\displaystyle n$ to have the same parity. Can you see why?
m and n have to have the same parity, is it because of the last equation $\displaystyle 2z=m^2+n^2$ (2z must be even)???
$\displaystyle x = m^2/2 - n^2/2 + mn$
$\displaystyle y = m^2/2 - n^2/2 - mn$
$\displaystyle z = m^2/2 + n^2/2$

But the theorem above requires r and s to be integers of opposite parity with r>s>0 and (r,s)=1. Do we need m and n to satisfy all of these requirements as well? (I think we need to ensure all these conditions are satsified for m and n, otherwise the theorem above won't even be applicable?)

Thanks for helping!

10. Originally Posted by kingwinner
m and n have to have the same parity, is it because of the last equation $\displaystyle 2z=m^2+n^2$ (2z must be even)???
$\displaystyle x = m^2/2 - n^2/2 + mn$
$\displaystyle y = m^2/2 - n^2/2 - mn$
$\displaystyle z = m^2/2 + n^2/2$

But the theorem above requires r and s to be integers of opposite parity with r>s>0 and (r,s)=1. Do we need m and n to satisfy all of these requirements as well? (I think we need to ensure all these conditions are satsified for m and n, otherwise the theorem above won't even be applicable?)

Thanks for helping!
Having opposite parity helps ensure primitive Pythagorean triples. There are other Pythagorean triples we can use though...

11. Originally Posted by chiph588@
Having opposite parity helps ensure primitive Pythagorean triples. There are other Pythagorean triples we can use though...
The theorem above is for PRIMITIVE Pyth. triples. Is there another theorem that says x,y,z is a Pythagorean triple (may or may not be primitive, just ANY Pythagorean triple) IF AND ONLY IF x,y,z are of the form $\displaystyle x = r^2 - s^2, y = 2rs, z = r^2 + s^2$ where r>s>0 are integers??

Also, in our problem, when we solve for x,y,z, we get terms like $\displaystyle m^2/2$ and $\displaystyle n^2/2$, but are they well-defined? (i.e. are they necessarily integers?)
$\displaystyle x = m^2/2 - n^2/2 + mn$
$\displaystyle y = m^2/2 - n^2/2 - mn$
$\displaystyle z = m^2/2 + n^2/2$

12. (m,n)=1 implies primitive

(m,n)=/=1 implies not primitive.

The formula you have above is the only formula for pythagorean triples. I believe I've given you all the information you need to solve the problem.

13. Theorem: The positive primitive solutions of $\displaystyle x^2 + y^2 = z^2$ with y even are precisely $\displaystyle x = r^2 - s^2, y = 2rs, z = r^2 + s^2$, where r and s are arbitrary integers of opposite parity with r>s>0 and gcd(r,s)=1.

The above theorem characterizes all "PRIMITIVE Pythagorean triples", but what is the statement of the theorem that characterizes ALL "Pythagorean triples" (not necessarily primitive)?

(i) The positive solutions of $\displaystyle x^2 + y^2 = z^2$ with y even are precisely $\displaystyle x = (r^2 - s^2)d, y = (2rs)d, z = (r^2 + s^2)d$, where d is any natural number, r and s are arbitrary integers of opposite parity with r>s>0 and gcd(r,s)=1.

(ii) The positive solutions of $\displaystyle x^2 + y^2 = z^2$ with y even are precisely $\displaystyle x = r^2 - s^2, y = 2rs, z = r^2 + s^2$, where r and s are arbitrary integers with r>s>0.

Which one is correct??

14. Go with (i)

15. Originally Posted by chiph588@
Go with (i)
I also think that (i) is correct.

Is (ii) wrong?
i.e. not all Pythagorean triple is of the form $\displaystyle x = r^2 - s^2, y = 2rs, z = r^2 + s^2$, r>s>0?
For example, 6,8,10 is a Pythagorean triple, but I think it cannot be written in the above form?

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