$3y^2=u^2-v^2,y^2=v^2-w^2$ can be easily rearranged to $\{a,b\},\{a,2b\}$ being such that they are positive and the sum of the squares of each pair is another square. We proceed by Infinite Descent, supposing $a+b$ is minimum, from which follow easily $(a,b)=1$ and $a$ odd, $b$ even. Therefore, by standard parametrization of pythagorean triples, $a=x^2-y^2=z^2-w^2, b=2xy=zw$. Therefore, $2x=m_1n_1,y=m_2n_2,z=m_1n_2,w=m_2n_1$ for some positive integers $m_1,m_2,n_1,n_2$. So we get $n_1^2(m_1^2+4m_2^2)=4n_2^2(m_1^2+m_2^2)$. It's easy to see that $p \mid m_1^2+m_2^2,m_1^2+4m_2^2 \Rightarrow p \mid m_1,m_2$. But if they share a factor, then $p^2 \mid a,b$, whcih cannot happen. Hence, $m_1^2+m_2^2,m_1^2+4m_2^2$ are squares. But this is a contradiction, since $a+b =z^2-w^2+2xy \geq z+w+x+\frac{y}2 > \frac12(z+w+2x+y)=\frac{(n_1+n_2)(m_1+m_2)}{2} \geq m_1+m_2$

$3y^2=u^2-v^2,y^2=v^2-w^2$ can be easily rearranged to $\{a,b\},\{a,2b\}$ being such that they are positive and the sum of the squares of each pair is another square. We proceed by Infinite Descent, supposing $a+b$ is minimum, from which follow easily $(a,b)=1$ and $a$ odd, $b$ even. Therefore, by standard parametrization of pythagorean triples, $a=x^2-y^2=z^2-w^2, b=2xy=zw$. Therefore, $2x=m_1n_1,y=m_2n_2,z=m_1n_2,w=m_2n_1$ for some positive integers $m_1,m_2,n_1,n_2$. So we get $n_1^2(m_1^2+4m_2^2)=4n_2^2(m_1^2+m_2^2)$. It's easy to see that $p \mid m_1^2+m_2^2,m_1^2+4m_2^2 \Rightarrow p \mid m_1,m_2$. But if they share a factor, then $p^2 \mid a,b$, whcih cannot happen. Hence, $m_1^2+m_2^2,m_1^2+4m_2^2$ are squares. But this is a contradiction, since $a+b =z^2-w^2+2xy \geq z+w+x+\frac{y}2 > \frac12(z+w+2x+y)=\frac{(n_1+n_2)(m_1+m_2)}{2} \geq m_1+m_2$

We firstly set up a lemma. (Based on AMM 10622) Lemma. Let $a<b<c$ are positive integers forming an arithmetic progressing such that $ab+1,bc+1,ca+1$ are all perfect squares. Then there exists $x,y \in \mathbb{Z}_{+}$ such that $a=2y-x,b=2y,c=2y+x,x^2-3y^2=1$. Proof. By using $a+c=2b$, we could obviously see that $4(ab+1)(bc+1)(ca+1)=(2abc+3b)^2+4ac+4-b^2$. Note that $4ac+4-b^2<4abc+6b+1=(2abc+3b+1)^2-(2abc+3b)^2$. And $4ac+4-b^2>-4abc-6b+1=(2abc+3b-1)^2-(2abc+3b)^2$. Which shows $(2abc+3b-1)^2<(2abc+3b)^2+4ac+4-b^2<(2abc+3b+1)^2$. Since $(2abc+3b)^2+4ac+4-b^2=4(ab+1)(bc+1)(ca+1)$ is a perfect square, we have $4ac+4-b^2=0$. Then $b$ is even, and define $b=2y$ and $x=c-b=b-a$, we could finish the proof of the lemma. WLOG $a<b<c<d$. Now we could use the lemma on both $a,b,c$ and $b,c,d$, which shows there exists positive integers $x,y,u,v$ such that $x^2-3y^2=u^2-3v^2=1$, and $a=2y-x,b=2y,c=2y+x,b=2u-v,c=2u,d=2u+v$. We could see that $x,v$ are even. Let $x=2z,v=2w$, then $y=u-w, v=y+z \Rightarrow z=3w-u$. Then use $u,w$ to replace $x,y,v$, we have $4(3w-u)^2-3(u-w)^2=u^2-12w^2=1$. Which is equivalent to $u^2-18uw+9w^2=u^2-12w^2=1$, which is obviously impossible.