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Let $x,y,z \in \mathbb{N}$ with $x,y$ relatively prime, $x$ even and $x^{2}+y^{2}=z^{2}$. Show that there are infinitely many $(x,y,z)$ triplets which satisfy these conditions.

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You should be able to use the answers in your other question to prove your claim. In particular, to borrow notation from your other question, there are an infinite number of relatively prime pairs $(u,v)$ with $u > v$; thus... –  J. M. Nov 2 '11 at 12:02
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@J.M. Duplicate maybe? –  user13838 Nov 2 '11 at 12:08
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You don't really need to know that $(u^2-v^2,2uv,u^2+v^2)$ gives all solutions to know that it gives infinitely many solutions. –  Thomas Andrews Nov 2 '11 at 13:29
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For example, given any $n$, $(x,y,z)=(8n,16n^2-1,16n^2+1)$ is a solution with $x$ even and $y,z$ relatively prime. –  Thomas Andrews Nov 2 '11 at 13:36
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@EmmadKareem Well, I could have chosen the simpler $(4n,4n^2-1,4n^2+1)$. Somehow, thought I needed an extra power of 2. –  Thomas Andrews Nov 2 '11 at 15:29

3 Answers 3

up vote 4 down vote accepted

It is easy to notice that $2s+1=(s+1)^2-s^2$, i.e., every odd number is a difference of two squares. Now if I take any odd square, I can get a Pythagorean triangle from this. There are infinitely many odd squares.

If I want to write this explicitly, I can put $(2k+1)^2=4k^2+4k+1=2[2k(k+1)]+1$, so I just plug $s=2k(k+1)$ into above formula and I get $$(2k+1)^2=(2k^2+2k+1)^2-(2k^2+2k)^2$$ i.e., $$(2k+1)^2+(2k^2+2k)^2=(2k^2+2k+1)^2.$$

Of course I do not obtain all Pythagorean triples in this way. This are only triples where the difference of the hypotenuse and one of the legs is one. (Note that all such triples are primitive since $\gcd(s,s+1)=1$.)

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Nice answer... You should also mention, even if it is obvious, that all these triples are rel prime :) –  N. S. Nov 2 '11 at 16:05

Euclid's formula for generating Pythagorean triples states that:

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

Let's show that if $\gcd(x,y)=1 \Rightarrow \gcd(m,n)=1$

suppose that $\gcd(m,n)>1$ , so we may write $m=k\cdot n$ , where $k>1 , n>1$ ,therefore:

$x=2kn^2 , y=n^2(k^2-1) \Rightarrow n^2 \mid x$ and $ n^2 \mid y \Rightarrow \gcd(x,y)>1$ which is false since we know

that $\gcd(x,y)=1$ . This means that our assumption $\gcd(m,n)>1$ is incorrect so $\gcd(m,n)=1$

Now let's show that $m-n$ is an odd number:

Since: $\gcd(x,y)=1$ and $x$ is even $\Rightarrow$ $m^2-n^2=(m-n)(m+n)$ must be an odd number,

therefore $m-n$ is odd number.

So we have this two conditions satisfied :

$\gcd (m,n)=1$ and $m-n$ is odd number

So we may conclude that $(x,y,z)$ is primitive Pythagorean triple and if you check a list of elementary properties of primitive Pythagorean triples you will find statement that there exist infinitely many primitive Pythagorean triples whose hypotenuses are squares of natural numbers so there are infinitely many $(x,y,z)$ triples.

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There are an infinite amount of u,v relative prime pairs with u>v , thus there are also infinite amount of numbers satisfying $\frac{1}{2}(z+y)\frac{1}{2}(z-y) = (\frac{x}{2})^{2}$ and that means there are infinite solutions of the form (x,y,z) for x even , x,y relatively prime. ? ? ?

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