# How can we prove that Gaussian integers are algebraic integers? [closed]

I heard in a lecture that every gaussian integer is in fact a root of a monic 2nd degree polynomial, it has been a day now and I can't figure this out.

• Hint: Multiply by the conjugate – Mummy the turkey Jan 27 at 19:29
• But that wouldn't be a polynomial, can you elaborate on that please. – Ahmed Omari Jan 27 at 19:31

It's a general fact that the sum of two algebraic numbers/integers is also one and clearly, $$bi = b\sqrt{-1}$$ and $$a$$ are algebraic integers so $$a+bi$$ is one too.
But in this example, we can be more explicit: $$a+bi$$ is the root of the polynomial $$x^2 - 2ax + (a^2+b^2) = 0$$.
In the second-degree polynomial $$f(x) = \Big( x-(a+bi)\Big)\Big( x - (a-bi)\Big),$$ the imaginary parts cancel out and the coefficients are then real integers.
• Here's a slightly different way of looking at it. Suppose we want to solve the quadratic equation $(x-a)^2+b^2=0$ where $a$ and $b$ are real integers. \begin{align} & (x-a)^2 + b^2 = 0. \\ {} \\ & (x-a)^2 = -b^2 \\ {} \\ & x-a = \pm bi \\ {} \\ & x = a\pm bi. \end{align} Every Gaussian integer is the solution of a quadratic equation of this kind. – Michael Hardy Jan 28 at 18:18