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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 Oct 8 '17 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 Oct 8 '17 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 Oct 8 '17 at 23:28
  • $\begingroup$ And also that $\gcd(4n^2, n^4+4)=1$? $\endgroup$ – Grouper Oct 8 '17 at 23:32
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    $\begingroup$ $$(4t^4-s^4)^2+(4t^2s^2)^2=(4t^4+s^4)^2$$ $\endgroup$ – Piquito Oct 9 '17 at 1:59
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All Pythagorean triples can be generated by the formula: $$A=(m^2-n^2)k\qquad B=2mnk\qquad C=(m^2+n^2)k$$

For a triple to be primitive, it is necessary but not sufficient that $k=1$. This leaves $B=2mn$. The smallest $m$ is $2$; $m,n$ must be of opposite parity; $m>n$, meaning 2mn must be a multiple of $4$

Every side-B is a multiple of $4$, e.g. $\quad (3,4,5)\quad (15,8,17)\quad (5,12,13)\quad (35,16,37)\quad ...$

Among these are all powers of $2$ that are perfect squares so, there being infinite combinations of $(m,n)$, there are an infinite number of triples where $\sqrt{B}=\lfloor\sqrt{B}\rfloor.$

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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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