# Infinite number of primitive Pythagorean triples

• Mar 27th 2010, 09:45 AM
kingwinner
Infinite number of primitive Pythagorean triples
Prove that there exist an infinite number of primitive Pythagorean triples x,y, and z with y even such that y is a perfect cube.

I can write down the definition, but I have no idea how prove this...

Any help is appreciated!

[also under discussion in math links forum]
• Mar 27th 2010, 09:55 AM
tonio
Quote:

Originally Posted by kingwinner
Prove that there exist an infinite number of primitive Pythagorean triples x,y, and z with y even such that y is a perfect cube.

I can write down the definition, but I have no idea how prove this...

Any help is appreciated!

Hint: if $\displaystyle a,b,c\in\mathbb{Z}\,\,\,then\,\,\,a^2+b^2=c^2\Long leftrightarrow a=m^2-n^2\,,\,b=2mn\,,\,c=m^2+n^2\,,\,\,m,n\in\mathbb{Z}$ .

The triple $\displaystyle (a,b,c)$ is called primitive if $\displaystyle gcd(m,n)=1$ and exactly one of them is even , so: can you find infinite pairs like $\displaystyle m,n$ as above? Well, there you have your infinite primitive pythagorean triples. (Cool)

Tonio
• Mar 27th 2010, 09:55 AM
chiph588@
Quote:

Originally Posted by kingwinner
Prove that there exist an infinite number of primitive Pythagorean triples x,y, and z with y even such that y is a perfect cube.

I can write down the definition, but I have no idea how prove this...

Any help is appreciated!

WLOG We know for some $\displaystyle m>n$

$\displaystyle x=m^2-n^2$
$\displaystyle y=2mn$
$\displaystyle z=m^2+n^2$

So let $\displaystyle m=4k^3$ and $\displaystyle n=1$.

Now observe $\displaystyle y=2\cdot4k^3 = (2k)^3$.
• Mar 28th 2010, 12:41 AM
kingwinner
Quote:

Originally Posted by tonio
Hint: if $\displaystyle a,b,c\in\mathbb{Z}\,\,\,then\,\,\,a^2+b^2=c^2\Long leftrightarrow a=m^2-n^2\,,\,b=2mn\,,\,c=m^2+n^2\,,\,\,m,n\in\mathbb{Z}$ .

The triple $\displaystyle (a,b,c)$ is called primitive if $\displaystyle gcd(m,n)=1$ and exactly one of them is even , so: can you find infinite pairs like $\displaystyle m,n$ as above? Well, there you have your infinite primitive pythagorean triples. (Cool)

Tonio

Yes, but the question requires also that y must be a perfect cube.
• Mar 28th 2010, 12:46 AM
kingwinner
Quote:

Originally Posted by chiph588@
WLOG We know for some $\displaystyle m>n$

$\displaystyle x=m^2-n^2$
$\displaystyle y=2mn$
$\displaystyle z=m^2+n^2$

So let $\displaystyle m=4k^3$ and $\displaystyle n=1$.

Now observe $\displaystyle y=2\cdot4k^3 = (2k)^3$.

Thanks, but can you explain how you came up with these answers and how can we prove that the resulting triples will all be "primitive" Pythag. triple and that y is a perfect cube?

I have seen the 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.

Thank you!
• Mar 28th 2010, 01:47 AM
tonio
Quote:

Originally Posted by kingwinner
Thanks, but can you explain how you came up with these answers and how can we prove that the resulting triples will all be "primitive" Pythag. triple and that y is a perfect cube?

I have seen the 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.

Thank you!

Well, you can choose ANY pair of coprime $\displaystyle m,n\in\mathbb{N}$ with say even m, so he chooses $\displaystyle m=4k^3\,,\,n=1$...it is obvious these pairs are coprime, right? By the way, he could as well choose $\displaystyle n=1\,,\,m=72k^3$

Tonio
• Mar 29th 2010, 12:30 AM
kingwinner
Quote:

Originally Posted by tonio
Well, you can choose ANY pair of coprime $\displaystyle m,n\in\mathbb{N}$ with say even m, so he chooses $\displaystyle m=4k^3\,,\,n=1$...it is obvious these pairs are coprime, right? By the way, he could as well choose $\displaystyle n=1\,,\,m=72k^3$

Tonio

Is there any restriction on k?

Also, are you sure that $\displaystyle m=72k^3$ would work? 144 is not a perfect cube...
• Mar 29th 2010, 05:01 AM
tonio
Quote:

Originally Posted by kingwinner
Is there any restriction on k?

Also, are you sure that $\displaystyle m=72k^3$ would work? 144 is not a perfect cube...

A mistake: it should have been $\displaystyle m=2^2\cdot 3^3k=108k$

Tonio
• Mar 29th 2010, 07:05 AM
kingwinner
In general, to ensure that the triple is "primitive", we need to ensure TWO things, i.e. (i) m and n must be of opposite parity AND (ii) gcd(m,n)=1, right?
• Mar 29th 2010, 07:22 AM
chiph588@
Quote:

Originally Posted by kingwinner
In general, to ensure that the triple is "primitive", we need to ensure TWO things, i.e. (i) m and n must be of opposite parity AND (ii) gcd(m,n)=1, right?

Right. Letting n=1 and m be a multiple of 4 always ensures this.
• Mar 29th 2010, 09:45 AM
kingwinner
Quote:

Originally Posted by chiph588@
Right. Letting n=1 and m be a multiple of 4 always ensures this.

Any reason why gcd(m,n)=1 alone is not enough to guarantee that the triple is "primitive"?
• Mar 29th 2010, 11:02 AM
chiph588@
Quote:

Originally Posted by kingwinner
Any reason why gcd(m,n)=1 alone is not enough to guarantee that the triple is "primitive"?

I think that's enough actually.
• Mar 29th 2010, 03:38 PM
kingwinner
Quote:

Originally Posted by chiph588@
I think that's enough actually.

I just went through the proof of the theorem fairly carefully, and I think we do need BOTH conditions to be satisfied in order to ensure gcd(x,y,z)=1.
• Mar 29th 2010, 03:39 PM
tonio
Quote:

Originally Posted by chiph588@
I think that's enough actually.

I don't think so: if both $\displaystyle m,n$ are odd then their squares' difference/sum are both even, rendering even integers in the triple...

Tonio