Let's say you have an equation $f(x)=0$ where $f:\mathbb R \to \mathbb R$ is composed of $$+,-,\times,\div,\sqrt{},n \in \mathbb Z$$ If $f$ has only finitely many solutions, are all the solutions algebraic?
I've tried (structural) induction. The cases $n$, $\div$ and $\times$ are easy. $+$ and $-$ are not.
I've been trying to reduce it to a polynomial equation. Problem is that doing $\sqrt{A}=B \implies A=B^2$ makes the RHS more complicated. If the RHS contains sums of square roots then this doesn't make progress. (Maybe I could do more work here?)
I don't see how to use the fact the algebraic numbers are closed under the above operations.
I've been trying to think of a counterexample.
Any radical expression exists in a tower of radical extensions of $\Bbb Q(x)$. For instance, consider
$$ T(x)=\sqrt{\sqrt{x+1}-\sqrt{x-1}}+\sqrt{x}-3.$$
Then $T$ exists in the top of the tower of radical extensions
$$ \begin{array}{c} \Bbb Q\left(\sqrt{x+1},\sqrt{x},\sqrt{x-1},\sqrt{\sqrt{x+1}-\sqrt{x-1}}\right) \\ | \\ \Bbb Q(\sqrt{x+1},\sqrt{x},\sqrt{x-1}) \\ | \\ \vdots \\ | \\ \Bbb Q(x) \end{array} $$
where $\Bbb Q(x)$ is the field of rational functions in $x$. Thus, $T$ is algebraic over $\Bbb Q(x)$, so it has a minimal polynomial, which in this case must have degree $\le 16$ since that's the degree of the extension up top. So $T$ is a root of an irreducible polynomial
$$ c_{16}(x)T^{16}+\cdots+c_1(x)T+c_0(x) \tag{$\ast$} $$
where all the $c_i$s are in $\Bbb Q[x]$ (by clearing denominators if necessary; the polynomial in $T$ needn't be monic), and $c_0(x)$ is not the zero polynomial in $x$ since the above polynomial in $T$ is irreducible. If $x$ is a root of the radical equation $T(x)=0$, then plugging it into $(\ast)$ yields $c_0(x)=0$, making $x$ the root of a polynomial equation - algebraic.