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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $$(4t^4-s^4)^2+(4t^2s^2)^2=(4t^4+s^4)^2$$ $\endgroup$ – Piquito Oct 9 '17 at 1:59

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