What are the solutions to x^2+1=0 in Z_3, Z_2 and C. Is the answer as simple as: no solutions in Z_3 and Z_2 and x=i,-i in C?
Hello, Duke!
$\displaystyle x^2+1\:\equiv\:0\text{ (mod 3)}$
We have: .$\displaystyle x^2 + 1 \:\equiv\:0\text{ (mod 3)}$
. . . . . . . . . . . $\displaystyle x^2 \:\equiv\:\text{-}1\text{ (mod 3)}$
. . . . . . . . . . . $\displaystyle x^2 \:\equiv\:2\text{ (mod 3)}\;\;\bf{[1]}$
We find that: .$\displaystyle \begin{Bmatrix} 0^2 &\equiv& 0 \text{ (mod 3)} \\ 1^2 &\equiv& 1\text{ (mod 3)} \\ 2^2 &\equiv& 1\text{ (mod 3)} \end{Bmatrix}$
Therefore, $\displaystyle \bf{[1]}$ has no solutions.
$\displaystyle x^2+1\:\equiv\:0\text{ (mod 2)}$
We have: .$\displaystyle x^2 + 1 \:\equiv\:0\text{ (mod 2)}$
. . . . . . . . . . . $\displaystyle x^2 \:\equiv\:\text{-}1\text{ (mod 2)}$
. . . . . . . . . . . $\displaystyle x^2 \:\equiv\:1\text{ (mod 2)}\;\;\bf{[2]}$
We find that: .$\displaystyle \begin{Bmatrix} 0^2 &=& 0 \text{ (mod 2)} \\ 1^2 &\equiv& 1\text{ (mod 2)} \end{Bmatrix}$
Therefore, $\displaystyle \bf{[2]}$ has one solution: $\displaystyle x \:\equiv\:1\text{ (mod 2)}$
$\displaystyle x^2 + 1 \:\equiv\: 0 \text{ (mod }C)$
We have: .$\displaystyle x^2 + 1 \:\equiv\:0\text{ (mod }C}$
. . . . . . . . . . . $\displaystyle x^2 \:\equiv\:\text{-}1\text{ (mod }C)$
. . . . . . . . . . . $\displaystyle x^2 \:\equiv\:C-1\text{ (mod }C)\;\;{\bf{[3]}$
$\displaystyle \bf{[3]}$ has solutions if $\displaystyle C \,=\,k^2+1$ for a positive integer $\displaystyle k.$
Are there any other solutions?
. . I don't know.