# $(2a)^2 +b^2=c^{4}a^2d^2 +4d$ 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 $4 \nmid d$.

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 $\equiv 1 \mod 4$.

• Thanks, i don't yet see how you got to the '-4 is square mod p' yet. Maybe basic stuff, but i'm not well informed on this. I'll think about it – Rutger Moody Sep 28 '16 at 20:58
• Since $a$ is invertible mod $p$, there is $c$ such that $b = a c$. Then $4 a^2 + b^2 \equiv 0$ becomes $c^2 \equiv -4$. – Robert Israel Sep 28 '16 at 21:40
• Thanks again! I now realise that quadratic reciprocity is sort of an all important theorem in these cases. I saw it come by earlier but didn't think it mattered. This certainly gives extra information on the structure. I still have hope i can reduce the possibilities further (that is: i don't think p=1 mod4 automatically means it must have solutions) I'll look into it. – Rutger Moody Sep 29 '16 at 8:23