Infinite number of primitive Pythagorean triples

**kingwinner** · March 27th 2010, 09:45 AM

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]

**tonio** · March 27th 2010, 09:55 AM

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

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

Tonio

**chiph588@** · March 27th 2010, 09:55 AM

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 $m>n$

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

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

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

**kingwinner** · March 28th 2010, 12:41 AM

Originally Posted by tonio

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

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

Tonio

Yes, but the question requires also that y must be a perfect cube.

**kingwinner** · March 28th 2010, 12:46 AM

Originally Posted by chiph588@

WLOG We know for some $m>n$

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

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

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

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 $x^2 + y^2 = z^2$ with y even are $x = r^2 - s^2, y=2rs, z = r^2 + s^2$ , where r and s are arbitrary integers of opposite parity with r>s>0 and (r,s)=1.

Thank you!

**tonio** · March 28th 2010, 01:47 AM

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 $x^2 + y^2 = z^2$ with y even are $x = r^2 - s^2, y=2rs, z = r^2 + s^2$ , 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 $m,n\in\mathbb{N}$ with say even m, so he chooses $m=4k^3\,,\,n=1$ ...it is obvious these pairs are coprime, right? By the way, he could as well choose $n=1\,,\,m=72k^3$

Tonio

**kingwinner** · March 29th 2010, 12:30 AM

Originally Posted by tonio

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

Tonio

Is there any restriction on k?

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

**tonio** · March 29th 2010, 05:01 AM

Originally Posted by kingwinner

Is there any restriction on k?

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

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

Tonio

**kingwinner** · March 29th 2010, 07:05 AM

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?

**chiph588@** · March 29th 2010, 07:22 AM

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.

**kingwinner** · March 29th 2010, 09:45 AM

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"?

**chiph588@** · March 29th 2010, 11:02 AM

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.

**kingwinner** · March 29th 2010, 03:38 PM

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.

**tonio** · March 29th 2010, 03:39 PM

Originally Posted by chiph588@

I think that's enough actually.

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

Tonio

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