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.
The Diophantine system $a^2 + b^2 = c^2$, $a^2 + nb^2 = d^2$ gives rise to the elliptic curve $$ E_n : y^2 = x (x+1) (x+n). $$ As is often the case for families of elliptic curves with a simple equation, this family has a long history, and still resists a complete answer even though we've made considerable progress since the 19th-century (and earlier) work reported in Dickson's History of the Theory of Numbers, Vol.2: Diophantine Analysis (see the section on "concordant forms"). It takes Cremona's program mwrank less than a minute to find that the sequence of integers $n > 1$ for which there are no solutions in nonzero $a,b,c,d$ begins
2, 3, 4, 5, 6, 8, 9, 12, 13, 14, 15, 16, 18, 19, 21, 25, 26, 28, 29, 32, 33, 35, 36, 37, 38, 39, 40, 43, 44, 46, 48, 51, 54, 55, 56, 62, 63, 64, 65, 66, 67, 69, 70, 73, 75, 78, 80, 81, 84, 87, 88, 89, 91, 95, 96, 98, … .
These are also the integers $n \in [2,100]$ for which $E_n$ has rank zero (the group of torsion points is isomorphic with $({\bf Z}/2{\bf Z}) \times ({\bf Z}/4{\bf Z})$ if $n$ is a square, and $({\bf Z}/2{\bf Z}) \times ({\bf Z}/2{\bf Z})$ otherwise). In each case this can be proved with a Fermat-style "$2$-descent". For each integer $n \in [2,100]$ not in that list, $E_n$ has rank $1$, except for $n=31$, $52$, $71$, $74$, $79$ for which $E_n$ has rank $2$. [The OEIS seems to contain neither the above sequence of $n$, nor the complementary sequence for which there are nonero solutions, nor the same sequences with $n$ replaced by $n-1$ which corresponds to $y^2 = x (x-1) (x+n)$.] For $n \leq 100$ the minimal solution gets as large as $$ (a,b,c,d) = (2873161, 2401080, 3744361, 22062761) $$ for $n=83$. Extending the computation to $n \leq 200$ finds the minimal solution $$ (a,b,c,d) = (13265620549, 6755532420, 14886702349, 80460628949), $$ for $n=138$, and also two cases ($n=124$ and $n=195$) where $E_n$ seems to have rank zero but mwrank cannot prove it.
To bring the system $a^2 + b^2 = c^2$, $a^2 + nb^2 = d^2$ into the "Weierstrass form" $y^2 = x (x+1) (x+n)$, start with the parametrization $(a:b:c) = (t^2-1 : 2t : t^2+1)$ of $a^2+b^2=c^2$ up to scaling, and find $a^2+nb^2 = t^4 + 2(2n-1) t^2 + 1$; if this is to be a square, we can write it as $(t^2-(2x+1))^2$ for some $x$ to obtain $(n+x) t^2 = x (x+1)$; given $x$ this has a solution $t$ iff $x (x+1) (x+n)$ is a square. The solutions with $b=0$ come from the subgroup $E_n[2]$ consisting of the "point at infinity" and the three $2$-torsion points at which $y=0$; if $n$ is a square, say $n=m^2$, then there are also solutions with $a=0$, and these come from $4$-torsion points such as $(x,y) = (m,m^2+m)$. From a non-torsion point $(x,y)$ we can recover $(a:b:c:d)$ by reversing the above procedure, or by doubling $(x,y)$ in the group law: the new point will have $x$-coordinate $(a/b)^2$, with each of the factors $x$, $x+1$, $x+n$ of $y^2$ being square separately.