The square of an algebraic number is also algebraic

Question

I'm trying to prove that if $x$ is an algebraic number then $x^2$ must also be.

It seems intuitive but I just can't find any kind of proof as I keep running into equations with fractional exponents that don't fit the definition. Any help please?

Answer 1

The number $x$ is algebraic if it is the root of a polynomial equation $p(x)=0$

Now set $y=x^2$ so that $p(x)=q(y)+xr(y)$ with $q(y), r(y)$ both polynomials.

Then $y$ is a root of $$\left(q(y)+xr(y)\right)\cdot\left(q(y)-xr(y)\right)=q^2(y)-x^2r^2(y)=q^2(y)-yr^2(y)=0$$which is a polynomial equation in $y$.