TITLE: $a^4 = x^2 + y^2 (6+8b)$ solutions $(x,y)$?
QUESTION [3 upvotes]: Given constants $a,b \in \mathbb{Z}$, can we find $(x,y)$ such that 
$a^4 = x^2 + y^2 16(6+8b)$
has a solution?

Alternatively: consider the ring $\mathbb{Z}[4\sqrt{-m}]$ with $m \equiv 6 \mod{8}$. I want to check if there are ideals of order $4$, i.e. $\mathfrak{a}^4$ is a principal ideal $(q)$. 
Taking norms 
$$a^4 = N(\mathfrak{a}^4) = N((q))=N(x+y4\sqrt{-m}) = x^2 + y^2 16m = x^2 + y^2 16(6+8b)$$

REPLY [0 votes]: You can parameterize the rational solutions by taking lines through the point $(a^2,0)$ and intersecting them with the curve. From this you can write down a formula for $x$ and $y$ in terms of the slopes of these lines, and from that you should be able to figure out which points are integral.
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