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

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.

• 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. – J.G. Dec 21 '18 at 20:43
• @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? – angryavian Dec 21 '18 at 21:58
• @angryavian I decided to multiply that by $2^2=4$ to avoid fractions. – J.G. Dec 21 '18 at 22:27
• @J.G. Oops, duh, sorry for not seeing that originally. – 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.

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