<s>[INST] $(2a)^2 +b^2=c^{4}a^2d^2 +4d$ integer solutions I'm trying to find possible values for the Diophantine equation:
$$
(2a)^2 +b^2=c^{4}a^2d^2  +4d \tag{1}\\
$$
With:
$$
\gcd(a,d)=1 \tag{2}
$$
Where $a,b,c,d $ all integers $ >0$. Specifically i'm interested in possible values for $d$.
For instance $d=1$ has solutions because for $d=1$ condition $(2)$ is automatically satisfied for every $a$. 
But $d=2$ has no solutions (i think, see proof below).
I'm hoping to prove that there are no solutions for $d \neq 1$.
I have two questions:
1) Does anyone know how to determine all possible values for $d$ in equation above?
2) I tried below to prove some statements myself ($d\neq 2,3 \nmid d, 4 \nmid d$ and $a,b,d$ have no common prime divisors $>2$). But i'm not shure they are correct. So if you can verify some of the statements below please let me know.
$(*1) \enspace d \neq 2$
$$
d=2 \implies \text{ everything must be divisible by } 4 \implies\\
a^2 +{{b^2}\over{4}}=c^{4}a^2  +2 \tag{3}\\
$$
$a$ must be odd because of $(2)$. If $c$ is odd then ${{b^2}\over{4}}$ must be even. But then ${{b^2}\over{4}}$ must be divisible by $4$ because it's a perfect square. Then if we take $(3) \pmod  4$ we get the contradiction : $1+0 \equiv 1+2    \pmod 4 $. So assume $c$ is even and ${{b^2}\over{4}}$ is odd. Rewriting $(3)$ shows that we can then isolate a term that must be divisible by $3$:   ${{b^2}\over{4}}=(c^{4}-1)a^2  +2=
 (c^{2}+1)(c-1)(c+1)a^2  +2 \implies {b^2\over{4}} \equiv 2  \pmod 3$  . Contradiction because $2$ is not in the quadratic residues $\pmod  3$ .

$(*2) \enspace 3 \nmid d$

Also $3$ can not divide $d$ : This would mean: $ 3 \mid (2a)^2 + b^2 \implies 3 \mid a^2 \land 3 \mid b^2 $ (see for example here). This is a contradiction with equation $(2)$.

$(*3) \enspace a,b,d $ 'coprime' for divisors $>2$

From $(1)$ and $(2)$ we see that any prime $p > 2$ that divides $d$ can not divide $a$ and therefore can also not divide $b$ because $d$ divides the right hand side of $(1)$ . So $\gcd(p,b)=1$ for prime $p>2$ with $p \mid d$.
Also a prime $p' > 2$ that divides $a$ can not divide $d$ and therefore can't divide $b$. So $\gcd(p',b)=1$.

$(*4) \enspace 4 \nmid d$

$$
d= \text{ even }  \implies a= \text{ odd and
 everything is divisible by } 4 \implies\\
a^2 +{{b^2}\over{4}}=c^4a^2{{d^{2}}\over{4}}  +d \\
a^2 +b'^2=c^4a^2d'^2  +2d' \\
d'= \text{ even } \implies \text{ right side divisible by } 4 \implies \text{ left side two odd squares }\\
\implies 1+1\equiv 0 \pmod 4 \implies \text{ contradiction }\\
$$
We conclude that $4 \nmid d$.
 [/INST] If $p$ is a prime dividing $d$, we need $4 a^2 + b^2 \equiv 0 \mod p$.  Since $a$ can't be divisible by $p$, that says $-4$ is a square mod $p$.  The primes with this property are $2$ and those $\equiv 1 \mod 4$.   
</s>