Reading GROSSWALD Emil "Representations of Integers as Sums of Squares" // GROSSWALD Emil "Representations of Integers as Sums of Squares" lesen

 Ein sehr schwer zu findendes Buch. Du kannst es dir hier ansehen.

A very hard-to-find book. You can view it here.

I have some of the books mentioned below in my Electronic Library.

The first encounter with the problem happened while reading SIERPIŃSKI W. Elementary Theory of Numbers (1988). On page 449 we have Problem 2(stated as Exercise) :

   " 2.   Find the complex integers  $x+y\imath$ which

 are representable as sums of the squares of two

 complex integers"

The exercise only has the answer, without any indication, not even bibliographical, of how to obtain it.

Read this solution yourself !!

A necessary condition is:  $y$  is an even number.

Then, for the number  $x+2y\imath$  we have :

$x+2y\imath\;=$  sums of the squares of two complex integers  $\Leftrightarrow$

$\Leftrightarrow$   NOT  both of  $\frac{x}{2}\;and\;y$  are odd integers

This formulation of the necessary and sufficient condition is inspired by NIVEN's article. The discussion is presented in more detail in GROSSWALD, pages 194-195.

        SIERPIŃSKI formulates, as I invited you to see, the necessary and sufficient condition in this way :

in  $x+y\imath\;\;y$  should be even and, in the case  $x=4t+2,\;\;y$  should be divisible by 4

Moreover, the corresponding writing is also presented, which no matter how much I searched, I could not find the way in which it was discovered.

ANSWER WS

$4t+4u\imath=[(t+1)+u\imath]^2+[u+(1-t)\imath]^2 \tag{1}$

$4t+(4u+2)\imath=[(t+u+1)+(u+1-t)\imath]^2+[(t-u)+(t+u)\imath]^2 \tag{2}$

$(2t+1)+2u\imath=[(t+1)+u\imath]^2+[u-t\imath]^2 \tag{3}$

$(4t+2)+4u\imath=[(t+u+1)+(u-t)\imath]^2+[(t-u+1)+(t+u)\imath]^2 \tag{4}$

                    Examples CiP

          Formulas (1)-(4) cover all possible cases of the numbers  $x+2y\imath$.  Indeed:

     -  when  $x \neq 4t+2$  we can have

                                                 $x=4t$,  when for  $y=2u$  we have (1)

 and for  $y=2u+1$  we have (2)                 

                                                  $x=4t+1\;\;or\;\;x=4t+3$, written together as

                                                            $x=2t+1$, and  $y$  can be any, having (3)

     -  when  $x=4t+2$  then necessarily  $y=2u$, and we have (4)