How to prove that expessions like $\sqrt{93+63\sqrt{85}} - \sqrt{143} \notin \Bbb{Z}$? The Problem:
There are multiple "rooty" equations that can be simplified to a whole number, for example:
$$\sqrt{19 + 6\sqrt{2}} - \sqrt{18} = 1$$
Because:
$$\sqrt{19 + 6\sqrt{2}} - \sqrt{18} = \sqrt{18+\sqrt{72} + 1} - \sqrt{18} = \\ = \sqrt{18+2\sqrt{18}+1} - \sqrt{18} = \sqrt{(\sqrt{18}+1)^2}-\sqrt{18} = \\ = \sqrt{18}+1-\sqrt{18} = 1$$
However, in the title, we have the expression of:
$$\sqrt{93+63\sqrt{85}} - \sqrt{143} \approx 14$$
And using a scientific calculator you indeed get $14$ as the answer. But using a high precision online calculator you get the true answer of:
$$\sqrt{93+63\sqrt{85}} - \sqrt{143} = 14.00000000005032...$$
The Question:
Is there a general way to prove that a "rooty" expression (like the one in the title) $\notin \Bbb{Z}$? Even if the expression is really close to a whole number, and even high precision calculators can't give you the correct answer?
Is there a general procedure or algorithm which tells you for sure that the number is or isn't $\in \Bbb{Z}$?
 A: In this example, if $\sqrt{93+63\sqrt{85}}-\sqrt{143}=t\in\Bbb Q$, then
$$93+63\sqrt{85}=(t+\sqrt{143})^2$$
but that implies that $\Bbb Q(\sqrt{85})=\Bbb Q(\sqrt{143})$ which is false.
So, not only is your expression not an integer, it's not rational either.
This sort of argument will work for most expressions like $\sqrt{a+\sqrt b}-\sqrt c$ but not for all possible "rooty" expressions.
A: A more general method that works for these type of problems is to first find the minimal polynomial of the root expression (this is the polynomial of which the expression is solution and it's also of the lowest possible degree). There is an algorithm for this and you can use wolframalpha. This only works for algebraic numbers however. 
For your example the minimal polynomial is $$x^8-944x^6-446946x^4-455778560x^2+112134658225$$
By the integral root theorem the divisors of $112134658225$ are the only possible integer solutions to this polynomial. The divisors are: $1,5,25,66973,334865,1674325,...$ and they get bigger and bigger. We also know that the root expression is solution to this polynomial and that it's close to $14$. However none of the divisors is anywhere close to $14$ so our root expression must be irrational solution. 
This method only works for when the leading coefficient is $1$ but can be generalised even further with the rational root theorem. 
I also recommend watching mathologers video here : https://www.youtube.com/watch?v=D6AFxJdJYW4 which more or less should answer some other questions you may have.
A: Let $K_0:=\Bbb{Q}$ and for $n\geq0$ define the field $K_{n+1}$ as
$$K_{n+1}:=K_n(\{\sqrt{x}:\ x\in K_n\})=K_n^{1/2},$$
and set $K:=\bigcup_{n\geq0}K_n$. Let $R\in K$ be a rooty expression, and let $n\in\Bbb{N}$ be such that $R\in K_n$. Let $x\in K_{n-1}$ be non-square. Then the map
$$\sigma_x:\ K_n\ \longrightarrow\ K_n:\ \sqrt{x}\ \longmapsto\ -\sqrt{x},$$
extends to a $K_{n-1}$-linear automorphism of $K_n$ that fixes $K_{n-1}$ pointwise. So if $\sigma_x(R)\neq R$ then $R\notin\Bbb{Z}$. On the other hand, if $\sigma_x(R)=R$ then $R\in K_{n-1}$, and we can repeat the same for some $x'\in K_{n-1}$ that is not a square. Eventually this will yield the minimal $n\in\Bbb{N}$ such that $R\in K_n$.
For example, for $R_1:=\sqrt{19 + 6\sqrt{2}} - \sqrt{18}$ we have $R_1\in K_2$ and we would like to check whether $x:=19+6\sqrt{2}$ is a square in $K_1$ (clearly $18$ is a square in $K_1$ because $18\in K_0$). If it is, then it is a square in $\Bbb{Q}(\sqrt{2})$, and
$$19+6\sqrt{2}=(a+b\sqrt{2})^2=(a^2+2b^2)+2ab\sqrt{2},$$
quickly shows that $(a,b)=(1,3)$ is a solution. Hence
$$R_1=\sqrt{19 + 6\sqrt{2}} - \sqrt{18}=1+3\sqrt{2}-3\sqrt{2}=1.$$
We can try the same for $R_2:=\sqrt{93+63\sqrt{85}} - \sqrt{143}\in K_2$. We would like to check whether $x:=93+63\sqrt{85}$ is a square in $K_1$. If it is, then it is a square in $\Bbb{Q}(\sqrt{85})$, but
$$93+63\sqrt{85}=(a+b\sqrt{85})^2=(a^2+85b^2)+2ab\sqrt{85},$$
is quickly checked to have no rational solutions. Then the automorphism of $K_2$ defined by
$$\varepsilon_x:\ K_2\ \longrightarrow\ K_2:
\ \sqrt{93+63\sqrt{85}}\ \longmapsto\ -\sqrt{93+63\sqrt{85}},$$
fixes $K_1$ pointwise, and hence
$$\sigma_x(R_2)=-\sqrt{93+63\sqrt{85}}+\sqrt{143}\neq R_2,$$
so $R_2\notin K_1$. In particular $R_2$ is not an integer.
