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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.

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    $\begingroup$ Hint: Multiply by the conjugate $\endgroup$ – Mummy the turkey Jan 27 at 19:29
  • $\begingroup$ But that wouldn't be a polynomial, can you elaborate on that please. $\endgroup$ – Ahmed Omari Jan 27 at 19:31
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There are a few things to say here:

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 $x^2 - 2ax + (a^2+b^2) = 0$.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Michael Hardy Jan 28 at 18:18

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