$a^2 + b^2 = c^2$, $ab = 6d^2$, $(a,b) = 1$ has no solutions $(a,b,c,d) \in \mathbb N^4$ 
Prove that $$a^2 + b^2 = c^2$$
$$ab = 6d^2$$
$$\gcd(a,b) = 1$$
has no solutions $(a,b,c,d) \in \mathbb N^4$.

Put $a = p^2 - q^2, b = 2pq, c= p^2 + q^2$ where $(p,q) = 1$, and $p - q \equiv 1 \bmod 2$ (i.e., one is odd, the other is even.) Then,
$$pq(p+q)(p-q) = 3d^2$$
$\{p,q,p+q,p-q\}$ are pairwise coprime. I thank Thomas Andrews for his thoughts and help with a similar (now deleted) question, which helped me make progress on this problem.
The following cases arise:

*

*$p = 3x_1^2, q = x_2^2, p+q = x_3^2, p-q = x_4^2$.

$3x_1^2 + x_2^2 = x_3^2$ gives $x_2^2 = x_3^2 \bmod 3$. $3x_1^2 = x_4^2 + x_2^2$ gives $x_4^2 + x_2^2 = 0 \bmod 3$, forcing $x_2^2 = x_3^2 = x_4^2 = 0 \bmod 3$. So, $3 \mid q$, which contradicts $(p,q) = 1$.


*$p = x_1^2, q = 3x_2^2, p+q = x_3^2, p-q = x_4^2$.

In this case, we get $x_1^2 = x_3^2 = x_4^2 = 1 \bmod 3$. I haven't been able to finish this part.


*$p = x_1^2, q = x_2^2, p+q = 3x_3^2, p-q = x_4^2$.

We have $x_1^2 + x_2^2 = 3x_3^2 = 0 \bmod 3$, forcing $x_1^2 = x_2^2 = 0 \bmod 3$. This contradicts $(p,q) = 1$.


*$p = x_1^2, q = x_2^2, p+q = x_3^2, p-q = 3x_4^2$.

$x_1^2 + x_2^2 = x_3^2 \bmod 3$, and $x_1^2 = x_2^2 \bmod 3$. Since $x_3^2 = 0$ or $1 \bmod 3$, we have $x_1^2 = x_2^2 = x_3^2 = 0 \bmod 3$, contradicting $(p,q) = 1$.
I need help with case ($2$) only, assuming the other cases are correctly done. Thank you!
 A: Added, September 13:   in the book by Sally, Chapter 2 is called Rational Right Triangles  and the Congruent Number Problem, pages47-122.  Section 8 is pages 96-101. Theorem 8.8 is that 3 is not a congruent number, proof according to Genocchio (1855), pages 99-101. I see, Exercise 8.10  on page 101 is to prove that any prime $p \equiv 3 \pmod 8$  is not a congruent number. It uses exercise 8.9, that when prime $p \equiv 3 \pmod 8,$   the ideal generated by $p$ in $ \mathbb Z [\sqrt 2 ] $  is a prime ideal.
your case 2  has $$x_1^2 + 3 x_2^2  = x_3^2   $$   while $$x_1^2 - 3 x_2^2  = x_4^2   $$
There are no such  $(x_1, x_2)$  integers with both nonzero.
The phrase is "$3$  is not a congruent number"
https://en.wikipedia.org/wiki/Congruent_number
https://oeis.org/A003273
For example, Genocchi proved  that no prime number $p \equiv 3 \pmod 8$ is a congruent number.   Later he invented dumplings with dough  composed of a simple combination of wheat flour, egg, salt, and potato.
Worth learning how to prove  $3$  by infinite descent. Bastien listed the congruent numbers up to 100  in a brief note 1915 , volume 22  of L'intermediaire des Math, pages 231-232. title Nombres Congruents.
Alright, Keith Conrad says that Fermat proved $2$ and $3$ not congruent.
Modern treatment:  pages 96-101  in Roots to Research   by Sally,
preview:
https://www.google.com/books/edition/Roots_to_Research/nHxBw-WlECUC?hl=en&gbpv=1&printsec=frontcover
A: For case 2, we could use the trick of Fermat's Infinite Descent.
Observe that $$\left(\dfrac{x_3+x_4}{2}\right)^2+\left(\dfrac{x_3-x_4}{2}\right)^2=x_1^2$$, and $3|x_3-x_4$. So we let $\dfrac{x_3+x_4}{2}=m^2-n^2, x_3-x_4=2mn, x_1=m^2+n^2$, where $m,n$ are positive integers and $(m,n)=1$. Then we have $x_3=m^2+2mn-n^2$. On the other hand, we have $x_3^2-x_1^2=3x_2^2$, which becomes
$$4mn(m^2-n^2)=3x_2^2$$
So $x_2$ is even, let it be $2l$. Then the equation becomes
$$mn(m+n)(m-n)=l^2$$
Now we see that $(p,q,d),(m,n,l)$ are solutions of $xy(x+y)(x-y)=3z^2$. If we assume $(p,q,d)$ is the smallest solution (in terms of the sum), which clearly should exist, then we will produce $(m,n,l)$ which is even smaller, contradiction.
