Finding $n$ satisfying that there is no set $(a,b,c,d)$ such that $a^2+b^2=c^2$ and $a^2+nb^2=d^2$

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.

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.

$$a=t^2-1$$

$$b=2t$$

$$c=t^2+1$$

$$w=3t^2-1$$