Show that it doesn't exist any of natural number $ n = 4m + 3$ that $ n= x^2+y^2 $ for any natural x and y Show that it doesn't exist any of natural number  $ n = 4m + 3$ that $  n= x^2+y^2 $ for any natural x and y
Show that every prime number in form $ p=4m+1 $ could be showed as $ p = x^2+y^2$ (x and y are natural)
I checked it and it

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*It can be: $$x=2n+1 \text{ and } y=2k+1 \ \text{ OR } \ x=2n \text{ and } y=2k \ \text{ OR } x=2n \text{ and } y=2k+1 \text{ OR } x=2n+1 \text{ and } y=2k$$
You can check each case and you will see that it cannot be $x^2+y^2=4m+3$
For example,at the first case you will get:
$$x^2+y^2=(2n+1)^2+(2k+1)^2=4n^2+4n+1+4k^2+4k+1=4(n^2+n+k^2+k)+2=4m+2$$
