Perfect Pythagorean triangle

It is True.

Proof:

All Pythagorean primitive triplets $a, b, c$ (c = hypothenuse) are expressible as:

$a = p^2 - q^2; b = 2pq; c = p^2 + q^2$ where $p > q$ and $p,q \in \mathbb{N}$

and the area of such a right triangle equals $\frac{1}{2}ab = \frac{(p^2 - q^2)(2pq)}{2} = \boxed{(p^2 - q^2)(pq)}$ (i). If the area is to equal a perfect square, then we require $p^2 - q^2 = \alpha^{2m}$ and $pq = \beta^{2n}$ where $\alpha, \beta, m, n \in \mathbb{N}.$ and $\alpha \ne \beta$ . Let $q = \frac{\beta^{2n}}{p}$ and substitute this according to:

$p^2 - q^2 = p^2 - (\frac{\beta^{2n}}{p})^{2} = \alpha^{2m} \Rightarrow p^4 - \alpha^{2m}p^2 - \beta^{4n} = 0.$

Solving this quadratic for $p^2$ now gives:

$p^2 = \frac{\alpha^{2m} \pm \sqrt{\alpha^{4m} - 4(1)(-\beta^{4n})}}{2} = \frac{\alpha^{2m} \pm \sqrt{\alpha^{4m} + 4\beta^{4n}}}{2}$

In order for $p^2$ to be an integer, we require the discriminant, $\alpha^{4m} + 4\beta^{4n}$ , to be a perfect square. We can actually rewrite this as:

$\alpha^{4m} + 4\beta^{4n} = \alpha^{4m} + 4\beta^{4n} + 4\alpha^{2m}\beta^{2n} - 4\alpha^{2m}\beta^{2n} = (\alpha^{4m} + 4\alpha^{2m}\beta^{2n} + 4\beta^{4n}) - 4\alpha^{2m}\beta^{2n} = \boxed{(\alpha^{2m} + 2\beta^{2n})^2 - (2\alpha^{m}\beta^{n})^2}$ (ii).

Our discriminant in (ii) is simply a difference of two squares, which is further simplified into:

$[(\alpha^{2m} + 2\beta^{2n}) + 2 \alpha^{m}\beta^{n}] \cdot [(\alpha^{2m} + 2\beta^{2n}) - 2 \alpha^{m}\beta^{n}]$ ;

or $[(\alpha^{2m} + 2 \alpha^{m}\beta^{n} + \beta^{2n}) + \beta^{2n}] \cdot [(\alpha^{2m} - 2 \alpha^{m}\beta^{n} + \beta^{2n}) + \beta^{2n}]$ ;

or $\boxed{[(\alpha^{m} + \beta^{n})^2 + \beta^{2n}] \cdot [(\alpha^{m} - \beta^{n})^2 + \beta^{2n}]}$ (iii).

However, (iii) can only be a perfect square if either $\alpha$ or $\beta$ equals zero, which cannot occur since $\alpha, \beta \in \mathbb{N}.$ Conclusion: The area of a Pythagorean triangle can never be a perfect square.

$\mathbb{Q.E.D.}$