Find all Pythagorean triangles whose area is twice a perfect square.

Let $x$ and $y$ be sides of a Pythagorean triangle. Using different approaches give no result! For example $xy=4d^2$ and $x^2+y^2=D^2$ gives $(x+y)^2=D^2+8d^2$ so taking $d=1=D$ solves for $x+y=3$ but no choice of $x$ and $y$ for $x+y=3$ is Pythagorean triangle! Any idea? Thanks.

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    $\begingroup$ The trick is to first know the Pythagorean triples are the triples of the form $(2kab, k(a^2-b^2), k(a^2+b^2))$ with $a,\,b$ co-prime and $a-b,\,b,\,k\in\Bbb N$. Thus we seek $a,\,b,\,k$ so that $2k^2 ab(a^2-b^2)$ is a perfect square, or equivalently $2ab(a^2-b^2)$ is. $\endgroup$ – J.G. Dec 21 '18 at 20:43
  • $\begingroup$ @J.G. Sorry if I'm misunderstanding something: The area of your triangle is $k^2 ab (a^2 - b^2)$, and for this to be twice a perfect square we would want $\frac{1}{2} k^2 ab (a^2 - b^2)$ to be a perfect square, right? $\endgroup$ – angryavian Dec 21 '18 at 21:58
  • $\begingroup$ @angryavian I decided to multiply that by $2^2=4$ to avoid fractions. $\endgroup$ – J.G. Dec 21 '18 at 22:27
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    $\begingroup$ @J.G. Oops, duh, sorry for not seeing that originally. $\endgroup$ – angryavian Dec 21 '18 at 22:29

WLOG, the two legs are relatively prime; if they have a common factor, we can divide each side by it. That divides the area by the square of that common factor, and we have a new smaller triangle that also works.

So now, the product of the legs is twice the area, or four times a perfect square. That makes the product of the legs itself a perfect square, so each leg is a perfect square. We're looking for solutions to $a^4+b^4=c^2$.

That looks familiar - it's a close relative of the $n=4$ case of Fermat's last theorem, and I know I've seen a proof that there are no solutions before. I'll try to reconstruct it:

OK, false start - that got messy. Looking it up - it's on the Wikipedia page for infinite descent, Fermat's favorite trick. (The following proof is not quite the same; I used that page for reference, but didn't copy it exactly)

Suppose we have a solution to $a^4+b^4=c^2$ in relatively prime positive integers. WLOG, $a$ is odd. By the standard decomposition of Pythagorean triples, we can write $a^2=m^2-n^2, b^2=2mn, c=m^2+n^2$. Then, since $m$ and $n$ are relatively prime with $m$ odd and $n$ even, $m$ and $2n$ are both perfect squares. We get a new Pythagorean triple $a^2+n^2=m^2$, with the hypotenuse a perfect square and one leg twice a perfect square.
Now, take another step down. Set $m=p^2+q^2, n=2pq, a=p^2-q^2$. From $n$ being a perfect square, we get that $p$ and $q$ are themselves perfect squares. That gives us $(\sqrt{p})^4+(\sqrt{q})^4=(\sqrt{m})^2$, a new solution to the original equation. Since $\sqrt{m} \le m < m^2+n^2=c$, this new solution has a smaller hypotenuse than the one we started with. Continuing indefinitely, we will eventually run out of positive integers, and we get a contradiction. There are no solutions.
Or, with more familiar logic, we could follow the rules of well-ordering. If there is a solution, there must be one with least hypotenuse. Start there, and the contradiction comes immediately.

So, long story short, there are no Pythagorean triples with the triangle's area twice a perfect square.

  • $\begingroup$ Yes, I know that, and it's what I proved. The summary sentence was the error there; fixed now. $\endgroup$ – jmerry Dec 21 '18 at 22:38

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