TITLE: Does the equation $x^2+23y^2=2z^2$ have integer solutions?
QUESTION [10 upvotes]: I would like to show that the image of the norm map $\text N : \mathbb Z \left[\frac{1 + \sqrt{-23}}{2} \right] \to \mathbb Z$ does not include $2.$ I first thought that the norm map from $\mathbb Q(\sqrt{-23}) \to \mathbb Q$ does not have $2$ as its image either, so I tried to solve the Diophantine equation $$x^2 + 23y^2 = 2z^2$$ in integers.  

After taking congruences with respect to several integers, such as $2, 23, 4, 8$ and even $16,$ I still cannot say that this equation has no integer solutions. Then I found out that the map $\text{N}$ has a simpler expression and can be easily shown not to map to $2.$  
But I still want to know about the image of $\text N,$ and any help will be greatly appreciated, thanks in advance.

REPLY [5 votes]: Conclusion: it is likely that the  parametrization
$$ \color{blue}{ \left(7u^2 + 10uv -3 v^2 \right)^2 + 23  \left(u^2 -2uv - v^2  \right)^2 = 8 \left( 3u^2 +  uv + 2 v^2 \right)^2}  $$
gives all solutions to $x^2 + 23 y^2 = 8 z^2$ with $\gcd(x,y,z) = 1.$  There is a theorem that all primitive solutions can be found with a small finite number of such parametrizations. This version is better than the one at the end of my discussion, $z$ is discriminant $-23$ rather than $-92,$ that is why there is no longer a problem about $z \equiv 2 \pmod 4.$ Live and learn.
If you wish to see how this does by computer (I advise this), with $z = 3u^2 + uv + 2 v^2 \leq M$ for some upper bound $M,$ we can demand
$$ |u| \leq \sqrt {\frac{8M}{23}}, $$
$$ |v| \leq \sqrt {\frac{12M}{23}}. $$
Here is a raw search for primitive solutions with $z \leq 100.$
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    x    y    z
    3    1    2
    7    1    3
   25    1    9
    5    7   12
   15    7   13
   45    1   16
   59    7   24
    9   17   29
   81    7   31
   61   17   36
    1   23   39
  111    7   41
  105   17   47
  147    1   52
  149    7   54
   35   31   54
  135   31   71
   53   41   72
   63   41   73
  205   17   78
  163   31   78
   41   47   81
   73   49   87
  263    1   93
  259   17   96

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Next is by parametrization, where it was necessary to use absolute values to get a nice ordering. In order to reduce repetition I took only $u \geq 0.$ I notice how not all $\pm$ signs occur in $x,y$ here. To get all types, one would need four slightly different parametrizations, $(x,y)$ as written, then $(x,-y),$ $(-x,y),$ $(-x,-y).$
    z             |x|  |y|   z              x    y    z              u    v
    2              3    1    2             -3   -1    2              0    1
    2              3    1    2             -3   -1    2              0   -1
    3              7    1    3              7    1    3              1    0
    9             25    1    9            -25    1    9              1   -2
   12              5    7   12              5    7   12              2   -1
   13             15    7   13             15   -7   13              1    2
   16             45    1   16             45   -1   16              2    1
   24             59    7   24            -59    7   24              2   -3
   29              9   17   29             -9   17   29              3   -2
   31             81    7   31            -81   -7   31              1   -4
   36             61   17   36             61  -17   36              2    3
   39              1   23   39             -1  -23   39              1    4
   41            111    7   41            111   -7   41              3    2
   47            105   17   47           -105   17   47              3   -4
   52            147    1   52           -147   -1   52              2   -5
   54            149    7   54            149    7   54              4    1
   54             35   31   54            -35   31   54              4   -3
   71            135   31   71            135  -31   71              3    4
   72             53   41   72             53  -41   72              2    5
   73             63   41   73             63   41   73              5   -2
   78            163   31   78           -163   31   78              4   -5
   78            205   17   78            205  -17   78              4    3
   81             41   47   81            -41  -47   81              1    6
   87             73   49   87            -73   49   87              5   -4
   93            263    1   93            263    1   93              5    2
   96            259   17   96           -259  -17   96              2   -7
    z             |x|  |y|   z              x    y    z              u    v

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There is another way to talk about the behavior of the norm map. The binary quadratic forms of discriminant $-92$ are in three classes, reduced are $x^2 + 23 y^2,$ $3x^2 + 2xy + 8y^2,$ $3x^2 - 2xy + 8y^2.$ 
That is, under Gauss composition, the groups is cyclic order three and  $3x^2 + 2xy + 8y^2$ is a generator. It primitively represents $8.$ It also primitively represents infinitely many squares $z^2,$ and composition tells us that $8z^2$ is primitively represented by $x^2 + 23 y^2.$
The part about squares is best written in Dirichlet's version of composition, which was all we had until Bhargava gave other interpretations of Gauss. The form $3x^2 + 2xy + 8y^2$ is integrally equivalent ($SL_2 \mathbb Z$) to $3x^2 + 4 xy + 9 y^2.$
Dirichlet's methods tell us (page 49 in Cox, although the first edition has a typo, pasting corrected selection from second edition  ) that 
$$ \left( 3 u^2 + 4 uv + 9 v^2 \right)^2 = 9 U^2 + 4 UV + 3 V^2,   $$
where $$  U = u^2 - 3 v^2, \; \; \; V = 6uv + 4 v^2.  $$ 
Another arrangement of the symbols $a,a',B,C$ tells us
$$ \left( 9U^2 + 4 U V + 3 V^2 \right) \left(3 s^2 + 4 st + 9 t^2 \right) = 27 S^2 + 4 ST + T^2, $$
where
$$ S = Us-Vt, \; \; \; T = 9Ut + 3Vs +4Vt.  $$ Taking lower case $s=1,t=-1$ gives
$$ 8 \left(3 U^2 + 4 UV + 9 V^2 \right) = 27 S^2 + 4 ST + T^2, $$
where
$$ S = U +V, \; \; \; T = -9U + 3V - 4V = -9U -V.  $$
Finally
$$ (T+2S)^2 + 23 S^2 = 27 S^2 + 4 ST + T^2. $$
Put it together, we can, for example with $p$ prime, solve $x^2 + 23 y^2 = 2 p^2$ whenever $p = 3 u^2 + 2 uv + 8 v^2.$ This is possible for $p = 3$ and then for all $(23|p) = 1$ such that
$ w^3 - w + 1$ is irreducible $\pmod p.$ This last fact is not in Cox, it is class field theory from a 1991 article by Hudson and Williams. 
Maybe I should summarize: we get an explicit formula for $x,y$ in
$$ x^2 + 23 y^2 = 8  \left( 3 u^2 + 4 uv + 9 v^2 \right)^2  $$
given $u,v$ arbitrary integers. That is,
$$ \color{blue}{ \left(-7u^2 + 6uv + 25 v^2 \right)^2 + 23  \left(u^2 + 6uv + v^2  \right)^2 = 8 \left( 3u^2 + 4 uv + 9 v^2 \right)^2}.  $$
Now that I've corrected some errors of mine (I was out of practice) the final formula can be checked easily enough.