Square and cubic root of gaussian Integer

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I define the integer square root of an integer n as the largest number r with $r^2$ <= n. Is there an analogon for Gaussian integers?



With n = c+di I am looking for the number r = a+bi with maximal norm $q = a^2+b^2$ where $q^2 <= c^2 + d^2$.



Using $(a+bi)^2 = a^2 - b^2 + 2ab i$ and $a^2+b^2 = sqrt(c^2+d^2)$ i get the following three equations:



a) $a^2-b^2 = c$



b) $d = 2ab$



c) $a^2+b^2 = sqrt(c^2+d^2)$



Adding a) and c) I get



a') $2a^2 = c + sqrt(c^2+d^2)$



Solving for a and b I get something like



$a = sqrt frac(c + sqrt(c^2+d^2))2, b = fracd2a$ ,



where the root sign stands for the integer root of an integer in the above sense. But what about the norm of this integer expression?




radicals gaussian-integers




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    up vote
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    down vote

    favorite












    I define the integer square root of an integer n as the largest number r with $r^2$ <= n. Is there an analogon for Gaussian integers?



    With n = c+di I am looking for the number r = a+bi with maximal norm $q = a^2+b^2$ where $q^2 <= c^2 + d^2$.



    Using $(a+bi)^2 = a^2 - b^2 + 2ab i$ and $a^2+b^2 = sqrt(c^2+d^2)$ i get the following three equations:



    a) $a^2-b^2 = c$



    b) $d = 2ab$



    c) $a^2+b^2 = sqrt(c^2+d^2)$



    Adding a) and c) I get



    a') $2a^2 = c + sqrt(c^2+d^2)$



    Solving for a and b I get something like



    $a = sqrt frac(c + sqrt(c^2+d^2))2, b = fracd2a$ ,



    where the root sign stands for the integer root of an integer in the above sense. But what about the norm of this integer expression?




    radicals gaussian-integers




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I define the integer square root of an integer n as the largest number r with $r^2$ <= n. Is there an analogon for Gaussian integers?



      With n = c+di I am looking for the number r = a+bi with maximal norm $q = a^2+b^2$ where $q^2 <= c^2 + d^2$.



      Using $(a+bi)^2 = a^2 - b^2 + 2ab i$ and $a^2+b^2 = sqrt(c^2+d^2)$ i get the following three equations:



      a) $a^2-b^2 = c$



      b) $d = 2ab$



      c) $a^2+b^2 = sqrt(c^2+d^2)$



      Adding a) and c) I get



      a') $2a^2 = c + sqrt(c^2+d^2)$



      Solving for a and b I get something like



      $a = sqrt frac(c + sqrt(c^2+d^2))2, b = fracd2a$ ,



      where the root sign stands for the integer root of an integer in the above sense. But what about the norm of this integer expression?




      radicals gaussian-integers




      share|cite|improve this question











      I define the integer square root of an integer n as the largest number r with $r^2$ <= n. Is there an analogon for Gaussian integers?



      With n = c+di I am looking for the number r = a+bi with maximal norm $q = a^2+b^2$ where $q^2 <= c^2 + d^2$.



      Using $(a+bi)^2 = a^2 - b^2 + 2ab i$ and $a^2+b^2 = sqrt(c^2+d^2)$ i get the following three equations:



      a) $a^2-b^2 = c$



      b) $d = 2ab$



      c) $a^2+b^2 = sqrt(c^2+d^2)$



      Adding a) and c) I get



      a') $2a^2 = c + sqrt(c^2+d^2)$



      Solving for a and b I get something like



      $a = sqrt frac(c + sqrt(c^2+d^2))2, b = fracd2a$ ,



      where the root sign stands for the integer root of an integer in the above sense. But what about the norm of this integer expression?





      radicals gaussian-integers





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









      asked Jul 15 at 7:21









      Stephan Januar

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