If $(a^2+b^2) \mid (c^2+d^2)$ and $\gcd(a,b)=\gcd(c,d)=1$ and $\gcd(a,c)>1$, what can be said about the components?

Question

While working on a divisibility problem in integers $a,b,c,d$, with $\gcd(a,b)=\gcd(c,d)=1$, I've come up against the hypothetical condition $$ (a^2+b^2) \mid (c^2+d^2), \tag{$\star$} $$ where, also by hypothesis, $\gcd(a,c) > 1$.

Ultimately, I’m trying to prove that $a=\pm 1$ (which would contradict the hypothesis $\gcd(a,c)>1$ and thus solve the original problem), or $d=\pm 1$ (which would solve the original problem in a second, essentially unrelated, way).

Any hints on how to determine anything about any of $a,b,c,d$ and how they relate to each other would be appreciated.

EDIT: There are lots of solutions to ($\star$) satisfying all the conditions, e.g. $(a,b,c,d)=(7,3,91,19)$. So this isn't a quest to show there are no solutions. I just want to determine if there are any general relations between the components.

Answer 1

If $n|a^2+1$ then $n=s^2+t^2$ whit gcd(s,t)=1

Answer 2

For this purpose it is necessary to solve the equation:

$$qx^2+qy^2=v^2+z^2$$

If the next root whole $\sqrt{2q-1}$ then it will be possible for this simple case record decisions:

$$x=((2q-3)k^2+2qkt-q(3q-2)t^2\mp((2q-1)k^2-2qkt-q(q-2)t^2)\sqrt{2q-1})p^2+$$

$$+2(k-qt)(\sqrt{2q-1}\pm1)ps-(1\pm\sqrt{2q-1})s^2$$

$$.......$$

$$y=((2q-3)k^2-2(q-2)kt+q(q-2)t^2\mp((2q-1)k^2-2qkt-q(q-2)t^2)\sqrt{2q-1})p^2+$$

$$+2((k+(q-2)t)\sqrt{2q-1}\mp(qt-k))ps-(1\pm\sqrt{2q-1})s^2$$

$$.......$$

$$v=((2q-1)k^2-2q(2q-1)kt+q(3q-2)t^2\pm((2q-1)k^2-2qkt-q(q-2)t^2)\sqrt{2q-1})p^2-$$

$$-2(((2q-1)k-qt)\sqrt{2q-1}\mp(qt-k))ps+(1\pm\sqrt{2q-1})s^2$$

$$.......$$

$$z=((4q^2-4q-1)k^2-2q(2q-3)kt-q(2q^2-3q+2)t^2\mp$$$$\mp((2q-1)k^2-2qkt-q(q-2)t^2)\sqrt{2q-1})p^2+$$

$$+2(k-qt)(\sqrt{2q-1}\pm1)ps-((2q-1)\pm\sqrt{2q-1})s^2$$

$k,t,p,s - $ integers asked us.