Finding $n$ satisfying that there is no set $(a,b,c,d)$ such that $a^2+b^2=c^2$ and $a^2+nb^2=d^2$ Let us consider $n\ge 3\in\mathbb N$ which satisfy the following condition.
Condition : There exist no set of four non-zero integers $(a,b,c,d)$ such that
$$a^2+b^2=c^2\ \ \text{and}\ \ a^2+nb^2=d^2.$$
Then, here is my question.

Question : How can we find every such $n$?

Remark : Please note $n\ge 3$. This is because it is known that $n=2$ satisfies the condition. ($a^2+b^2=c^2,a^2+2b^2=d^2\Rightarrow c^2-b^2=a^2,c^2+b^2=d^2\Rightarrow c^4-b^4=(ad)^2$ and see, for example, here)
The followings are the examples of $n$ which do not satisfy the condition. 


*

*For $n=4k^2+5k+1\ (n=10,27,52,85,126,\cdots)$, take $(a,b,c,d)=(3,4,5,8k+5).$

*For $n=4k^2+3k\ (n=7,22,45,76,115,162,\cdots)$, take $(a,b,c,d)=(3,4,5,8k+3).$

*For $n=9k^2+10k+1\ (n=20,57,112,185,\cdots)$, take $(a,b,c,d)=(4,3,5,9k+5).$

*For $n=9k^2+8k\ (n=17,52,105,176,265,\cdots)$, take $(a,b,c,d)=(4,3,5,9k+4).$
We know that $(a,b,c)$ is a Pythagorean triple and we can see that $$\text{$(a,b,d)=\left((s^2-nt^2)u,2stu,(s^2+nt^2)u\right)\ $ satisfy $\ a^2+nb^2=d^2$}.$$ However, I don't have any good idea to find such $n$. Can anyone help?
Added : A user individ found that if there are integers $p,s,t$ such that 
$$(a,b,c,d,n)=(p-s,2t,p+s,\mp 2n+p+s\pm 2,(p\pm 1)(s\pm 1))\ \ \text{and}\ \ ps=t^2,$$
then the $n$ does not satisfy the condition.
However, this does not say anything about $n$ which satisfy the condition. We still don't know if each of $n=3,4,5,6,8$, for example, satisfies the condition.
 A: For the system of equations:  
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&a^2+qb^2=w^2\end{aligned}\right.$$  
If you can decompose the coefficient multipliers as follows:  $q=(p\pm1)(s\pm1)$  
Their work squares: $ps=t^2$  
Then decisions can be recorded.  $$a=p-s$$  $$b=2t$$  $$c=p+s$$  $$w=\mp2q+p+s\pm2$$
You can add another simple option.  If the ratio can be written as:  $$q=2t^2-1$$
Then decisions can be recorded.
$$a=t^2-1$$
$$b=2t$$
$$c=t^2+1$$
$$w=3t^2-1$$
A: What you are looking for are integers that are not concordant forms. To quote from the link, a concordant form is an integer triple $a,b,n$ such that,
$$a^2+b^2 = c^2$$
$$a^2+nb^2 = d^2$$
and integer $c,d$. As early as 1857, the list of solvable $n<100$ was given (though missing three terms, in blue):
$$\small n=1, 7, 10, 11, 17, 20, 22, 23, 24, 27, 30, 31, 34, 41, 42, 45, \color{blue}{47}, 49, 50, 52, \color{blue}{53}, 57, 58, 59, 60, 61, 68, 71, 72, 74, 76, 77, 79, 82, \color{blue}{83}, 85, 86, 90, 92, 93, 94, 97, 99, 100$$
For some bizarre reason, this is not yet in the OEIS. (Anyone care to submit it?)
Your question about the $n$ of non-concordant forms would simply be its complement, and given by Elkies' answer as $\small n = 2, 3, 4, 5, 6, 8, 9, 12, 13, 14, 15,\dots$
P.S. Incidentally, the concordant form with $n=52$ has a special use in equal sums of like powers. Let,
$$a^2+b^2 = c^2$$
$$a^2+52b^2 = d^2$$
with initial solution $a,b,c,d = 3,4,5,29$, and an infinite more. Then, 
$$(8b)^k + (5a-4b)^k + (-a-2d)^k +  (a-2d)^k + (-5a-4b)^k  + (-12b+4c)^k + (12b+4c)^k =\\ 
(4a+8b)^k + (3a-2d)^k + (-3a-2d)^k + (-4a+8b)^k + (-16b)^k + (a+4c)^k + (-a+4c)^k$$ 
for $k=1,2,4,6,8,10,$ found by J. Wroblewski and yours truly.
