How to prove that Φ² (golden ratio squared) is an algebraic number?

Question

The golden ratio Φ is algebraic since it is a root of the polynomial $x^2 - x - 1$

But how would you write and prove that Φ² is also algebraic?

Answer 1

If $\alpha$ is an algebraic number over $\mathbb{Q}$ then $\alpha^2$ is an algebraic number over $\mathbb{Q}$. Proof: let $p(x)$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and $p_{\text{even}}(x)=\frac{p(x)+p(-x)}{2}$, $p_{\text{odd}}(x)=\frac{p(x)-p(-x)}{2}$. We have $p_{\text{even}}(x)=r(x^2)$ and $p_{\text{odd}}(x)=x\cdot s(x^2)$, so $$ t(x) = r(x)^2-x\cdot s(x)^2 $$ is a polynomial with degree $\leq\deg p$ vanishing at $\alpha^2$, since $p(x)\mid t(x^2)$.

About the golden ratio, this procedure applied to the polynomial $x^2-x-1$ gives that $x^2-3x+1$ is a polynomial vanishing at $\varphi^2$.

Answer 2

$\Phi$ is a root of $f(x)=x^2 - x - 1$. Therefore, $\Phi^2 =\Phi+1$ and so $\Phi^2$ is a root of $f(x-1)=x^2 - 3 x + 1$.

More generally, $\Phi ^n = F_n \Phi + F_{n-1}$, where $F_n$ is the $n$-th Fibonacci number, and so $\Phi ^n$ is a root of $f\left(\dfrac{x - F_{n-1}}{F_n}\right)=\dfrac{x^2-L_n x +(-1)^n}{F_{n}^2}$, where $L_n$ is the $n$-th Lucas number, $L_n = F_{n-1}+F_{n+1}$. The value of the independent term follows from Cassini's identity. We can remove the denominator and so $\Phi ^n$ is a root of $x^2-L_n x +(-1)^n$.

This also follows by computing the trace and the determinant of the powers of the matrix $ \pmatrix{ 0 & 1 \\ 1& 1} $, which represents the map $ x \mapsto \Phi x$ in $\mathbb Q(\Phi)$: $$ \pmatrix{ 0 & 1 \\ 1& 1}^n = \pmatrix{ F_{n-1} & F_n \\ F_n & F_{n+1}} $$