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Prove that there exist infinitely many primitive Pythagorean triples $x, y, z$ whose even member $x$ is a perfect square. [Hint: consider the triple $4n^2, n^4-4, n^4+4$, where $n$ is an arbitraty odd integer.]

What I got:

Using the hint. $4n^2, n^4-4, n^4+4$ is a Phytagorean triple if $x^2+y^2=z^2$. Replacing and solving the equation it is clear that, $(4n^2)^2+(n^4-4)^2=(n^4+4)^2$ where $n$ is odd is indeed a Pythagorean triple with $x=4n^2=(2n^2)^2$ a perfect square.

Now I have to prove that $gcd(4n^2, n^4-4, n^4+4)=1$.

But I'm stuck here. I tried $gcd(n^4-4, n^4+4)=1$ without success. Any ideas?

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    $\begingroup$ Consider when $n$ is odd. Notice that $\gcd(n^4-4,n^4+4)=\gcd(n^4-4,8)=1$ when $n$ is odd (we used here that $\gcd(a,b)=\gcd(a,b-a)$) from which it follows that for all odd $n$, the $\gcd$ of all the three numbers is $1$. $\endgroup$
    – Prasun Biswas
    Commented Oct 8, 2017 at 21:05
  • $\begingroup$ As mentioned, I used the property $\gcd(a,b)=\gcd(a,b-a)$. If you want to see why this is true, show that $\gcd(a,b)\mid\gcd(a,b-a)$ and $\gcd(a,b-a)\mid\gcd(a,b)$ and conclude. $\endgroup$
    – Prasun Biswas
    Commented Oct 8, 2017 at 23:16
  • $\begingroup$ I didn´t mean to ask that last one. What I really want to ask is why it follows that the $\gcd$ of all the three numbers is $1$. Don´t I have to prove that $\gcd(4n^2, n^4-4)=1$ too? $\endgroup$
    – Grouper
    Commented Oct 8, 2017 at 23:28
  • $\begingroup$ And also that $\gcd(4n^2, n^4+4)=1$? $\endgroup$
    – Grouper
    Commented Oct 8, 2017 at 23:32
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    $\begingroup$ $$(4t^4-s^4)^2+(4t^2s^2)^2=(4t^4+s^4)^2$$ $\endgroup$
    – Piquito
    Commented Oct 9, 2017 at 1:59

2 Answers 2

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Instead of the formula proposed, we use the following formula which generates only the subset of Pythagorean triples (A,B,C) where $\,C-A=2.$

$A=4n^2-1\quad B=4n\quad C=4n^2+1$

We can see that, any time $\,n\,$ is a perfect square, then $\,B=4n,\,$ as the product of two squares, is also a perfect square.

$\therefore n\in\big\{1,4,9,16,\cdots\big\} \implies 4n\in\big\{4,16,36,64,\cdots\big\}\quad$ so there are an infinite number of triples where the even side is a perfect square.

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If $n$ is odd, $\gcd (n^4-4, n^4+4)=\gcd (n^4-4,8)=1$ by the Euclidean algorithm.

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