Factorizing $x^2+y^2+1$

Question

I got $x^2+y^2$ could factorized by $(x+yi)(x-yi)$

But Could we get factorization of $x^2+y^2+1$

I tried $(x+yi+i)(x-yi-i)$ but i couldn't guess it.

By FTA, It is possible but I couldn't gess it....

Answer 1

It is not possible to factor $x^2 + y^2 + 1$ into polynomials in $x$ and $y$, i.e. it is irreducible over $\mathbb C[x,y]$. Of course you could factor into polynomials in $x$ as $(x + \sqrt{-1-y^2})(x - \sqrt{-1-y^2})$, but these are not polynomials in $y$.

Answer 2

Look at the homogenization of your polynomial, viz. $x^2+y^2+z^2$. Finding a non-trivial factorisation $$x^2+y^2+z^2=(a'x+b'y+c'z)(a''x+b''y+c''z)$$ with complex coefficients is equivalent to decomposing the conic $x^2+y^2+z^2=0$ in tthe complex projective plane as the union of two lines. There is a general result that a conic $$\begin{bmatrix}x&y&z\end{bmatrix}\begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=[0]$$ in the complex projective plane is the union of two lines iff $$det\begin{bmatrix}a&h&g\\h&b&f\\g&f&c\end{bmatrix}=0.$$[Note that the proof of this result is highly dependent on the fact that the characteristic of the complex numbers is not 2,] For your polynomial, $a=b=c=1,h=g=f=0,$ so the detrminant is 1 and the conic does not decompose, i.e. the polynomial does not have a non-trivial factorisation.