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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.

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  • 3
    $\begingroup$ May I know the reason for down-vote to improve the next time? Thanks. $\endgroup$ – awllower Mar 1 '16 at 13:45
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    $\begingroup$ $7^2 + 23*1^2=2*6^2$. $\endgroup$ – Abstraction Mar 1 '16 at 13:53
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    $\begingroup$ And $3^{2}+23*1^{2}=2*4^{2}$. $\endgroup$ – Oscar Lanzi Mar 1 '16 at 19:23
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    $\begingroup$ And $6^2+23*2^2=2*8^2$. $\endgroup$ – Ali Caglayan Mar 1 '16 at 20:01
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    $\begingroup$ $ \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} $ $\endgroup$ – Will Jagy Mar 1 '16 at 21:12
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$x^2+23y^2=2z^2\iff x^2+(5y)^2=2(z^2+y^2)$.

Since the solutions of the equation $X^2+Y^2=2Z^2$ are given by the identity $$(a^2+2ab-b^2)^2+(a^2-2ab-b^2)^2=2(a^2+b^2)$$ we can try $(y,z)=(a,b)$ taking care of one of $a^2+2ab-b^2$ or $a^2-2ab-b^2$ be equal to $5b$.

Taking for example $(y,z)=(1,4)$ we get the solution $(x,y,z)=(3,1,4)$.

Thus the proposed equation have solutions (which can be parametrized but I stop here).

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  • $\begingroup$ Your first line just dawned on me! Thanks for this simple but beautiful identity! $\endgroup$ – awllower Mar 1 '16 at 15:45
  • $\begingroup$ nice....................+1 $\endgroup$ – Bhaskara-III Mar 1 '16 at 21:10
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    $\begingroup$ $ \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 primitive solutions $\endgroup$ – Will Jagy Mar 1 '16 at 21:36
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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

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

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

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

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.

enter image description here

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  • $\begingroup$ Thanks for this interpretation as quadratic forms and the solution following this line. $\endgroup$ – awllower Mar 2 '16 at 1:30
  • $\begingroup$ @awllower I may have been hasty. Were you asking how to describe the numbers represented by $x^2 + xy + 6 y^2?$ $\endgroup$ – Will Jagy Mar 2 '16 at 3:33
  • $\begingroup$ I don't think so: what I asked is already answered here. $\endgroup$ – awllower Mar 2 '16 at 3:40
  • $\begingroup$ @WilI Jagy: I read your quote Skolem to Albanese. I completely agree. The problem, with regard to the truth, and is almost always the case, is that who does not know these things well, prefers simplicity. Best regard. $\endgroup$ – Piquito Mar 2 '16 at 17:23
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To solve such problems, it is necessary to use the formula. Solving a Diophantine equation of the form $x^2 = ay^2 + byz + cz^2$ with the constants $a, b, c$ given and $x,y,z$ positive integers

The formula is written in a General form so I will have to solve the equation. For the equation.

$$23x^2+y_1^2=2z^2$$

Will do the replacement. $y_1=x+ay$ The equation will look like.

$$(23+a^2)x^2+2axy+y^2=2z^2$$

And we need to find this number $a$.

Using these formulae. And we will need to solve the Pell equation. $3$ use the formula. $$q=\sqrt{4a^2+4(23+a^2)(2-1)}=2\sqrt{23+2a^2}$$ This equation Pell. $q=10$ : $a=1$

So you need to solve the equation.

$$24x^2+2xy+y^2=2z^2$$

Using the formula the solution will be. Reducing a common divisor.

$$x=(40\mp54)p^2+4(1\mp5)ps\mp{s^2}$$

$$y=-(230\pm54)p^2+4(1\mp5)ps+(5\mp1)s^2$$

$$z=54(5\mp1)p^2+2(27\mp10)ps+(5\mp1)s^2$$

Remember that $y_1=x+y$ For the equation.

$$23x^2+y_1^2=2z^2$$

Finally the solution can be written as.

$$x=(40\mp54)p^2+4(1\mp5)ps\mp{s^2}$$

$$y_1=-(190\pm108)p^2+8(1\mp5)ps+(5\mp2)s^2$$

$$z=54(5\mp1)p^2+2(27\mp10)ps+(5\mp1)s^2$$

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  • $\begingroup$ Thanks for providing a parameterisation of the solutions. $\endgroup$ – awllower Mar 1 '16 at 16:17
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$$98**2 + 23*14**2 = 2*84**2$$

These triples work too.

1 23 78 3 1 4 5 7 24 6 2 8 7 1 6 9 3 12 9 17 58 10 14 48 12 4 16 14 2 12 15 5 20 15 7 26 15 21 72 18 6 24 20 28 96 21 3 18 21 7 28 24 8 32 25 1 18 27 9 36 28 4 24 30 10 40 30 14 52 33 11 44 35 5 30 36 12 48 39 13 52 42 6 36 42 14 56 45 1 32 45 15 60 45 21 78 48 16 64 49 7 42 50 2 36 51 17 68 54 18 72 56 8 48 57 19 76 59 7 48 60 20 80 61 17 72 63 9 54 63 21 84 66 22 88 69 23 92 70 10 60 72 24 96 75 3 54 77 11 66 81 7 62 84 12 72 90 2 64 91 13 78

Program dio.py

for k in range(100): for l in range(100): for m in range(100): if k*k + 23*l*l == 2*m*m: print(k,l,m)

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  • $\begingroup$ Thanks for the great help. $\endgroup$ – awllower Mar 1 '16 at 14:17
  • $\begingroup$ bad habit reward the result of a calculator $\endgroup$ – Piquito Mar 1 '16 at 14:45
  • $\begingroup$ I will paste in my brute-force effort so you can see it. $\endgroup$ – ncmathsadist Mar 1 '16 at 14:53
  • $\begingroup$ excuse me but I did not offend you. I thought overall mod $\endgroup$ – Piquito Mar 1 '16 at 14:56
  • $\begingroup$ $ \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 primitive solutions. $\endgroup$ – Will Jagy Mar 1 '16 at 21:24

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