prove that if there exist a pythogarian triangle whose area is the square of an integer that there exist $x,y,z \in \mathbb{N}$ such that:

$x^4-y^4 = z^2$

I tried to look at a pythogarian triple $(a,b,c)$ with this property, so we see $\frac{1}{2}ab = k^2$ for some $k \in \mathbb{N}.$

I see that $(x^2 + y^2)(x^2-y^2) = z^2$, so proving that both terms on the left are squares will help me solve the problem.

I tried using primitive Pythagorean triangles, which can be gained by dividing al integers by the gcd, so we can get a new pair (x,y,z) where, $x = 2pr$, $y = p^2-r^2$ and $z = p^2 + r^2$. because $pr(p^2 -r^2)= k^2$, we see that $p^2 - r^2$ is a square, but how do i know that $p^2 + q^2$ is square?

any hints woud be much appreciated :)



If there is such a triple $(a,b,c)$, there is such a triple which in addition is primitive. By the representation theorem for primitive triples, there exist integers $s$ and $t$, relatively prime and of opposite parity, such that $$a=s^2-t^2 \quad \text{and} \quad b=2st.$$ We are told that $st(s^2-t^2)$ is a perfect square. Since $s$, $t$, and $s^2-t^2$ are pairwise relatively prime, it follows that $s=x^2$, $t=y^2$ and $s^2-t^2=z^2$ for some integers $x,y,z$.

Thus $x^4-y^4=z^2$.


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