How to prove that expessions like $\sqrt{93+63\sqrt{85}} - \sqrt{143} \notin \Bbb{Z}$?

Question

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 $\in \Bbb{Z}$?

Can you clarify what constitutes a "rooty" expression? Do you allow $n$-th roots for all positive integers $n$, or just square roots? Do you allow division as well? — Servaes, Mar 31 '19 at 10:50
Rooty would mean any expression containing $n$th roots, $n$th powers, and the 4 basic operations of $+, -, *, \backslash$. — Daniel P, Mar 31 '19 at 10:53
However, it would be nice to even get a way of solving the problem for more general expressions, like ones only containing square roots, $+, - ,$ and $*$ — Daniel P, Mar 31 '19 at 10:55
Easier: by my simple denesting rule $,\sqrt{\color{#0a0}{6\sqrt{2}!+!19}}= (\color{#0a0}{6\sqrt{2}!+!19}!-!17)/2 = 3\sqrt 2+1,,$ so subtracting $,\sqrt{18}=3\sqrt 2,$ yields $1.\ \ $ — Bill Dubuque, Jun 16 '24 at 07:57

Angina Seng · Answer 1 · 2019-03-31T11:21:22.487

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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.

edited Mar 31 '19 at 11:21

answered Mar 31 '19 at 11:03

Angina Seng

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The question was with $\sqrt{143}$, not $\sqrt{14}$ – Peter Foreman Mar 31 '19 at 11:12
@PeterForeman mutatis mutandis – Angina Seng Mar 31 '19 at 11:21
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Your argument would work for most cases, however, there might be expressions of the form $\sqrt{a+b\sqrt{x}} - d\sqrt{x}$, for which you'd get $a+b\sqrt{x} = (t+d\sqrt{x})^2$, and you cannot follow the same argument of $\Bbb{Q}(\sqrt{x}) \ne \Bbb{Q}(\sqrt{x})$. – Daniel P Mar 31 '19 at 11:27

score 3 · Answer 2 · answered Mar 31 '19 at 11:10

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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.

answered Mar 31 '19 at 11:10

DreaDk

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What if the divisors are close to $14$ though?
And I've watched the Mathologer video, but those methods only work when the rooty expression is exactly a whole number, and I wan't to prove that it isn't.

Daniel P

Mar 31 '19 at 11:30

@DanielP You simply plug in all values allowed by the rational root theorem; if none of them is a zero, then the minimal polynomial has no rational roots, and hence your rooty expression is not rational. – Servaes Mar 31 '19 at 11:46

I've read up on the rational root theorem in the mean time, and it's a pretty smart approach. However, what happens if the minimal polynomial happens to have the roots $14$ and $14.00000000005032...$? – Daniel P Mar 31 '19 at 11:55

I don't think you will ever get such case in reality. You could technically construct such polynomial but if an expression seems to be converging to whole number and the root of the minimal polynomial is that number , it's almost certain that the root expression is then equal to that number. – DreaDk Mar 31 '19 at 12:09

Also check root separation of polynomials. There is usually a lower bound on how close two distinct roots of polynomial can be. Knowing the minimal polynomial you could figure out if having roots that close to each other is even possible and then conclude whether your expression is rational or not depending on that. – DreaDk Mar 31 '19 at 12:22

score 1 · Accepted Answer · answered Mar 31 '19 at 11:26

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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.

answered Mar 31 '19 at 11:26

Servaes

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This is great, however, it only works for most cases. Consider an expression of the form $\sqrt{a+b\sqrt{x}} - d\sqrt{x}$, (that is close to an integer). Proving it for an expression like this would be tough, since you can't disprove that $\Bbb{Q}(\sqrt{x}) = \Bbb{Q}(\sqrt{x})$. – Daniel P Mar 31 '19 at 11:36
@DanielP I don't understand; either $a+b\sqrt{x}$ is a square in $\Bbb{Q}(\sqrt{x})$, in which case $\sqrt{a+b\sqrt{x}}-d\sqrt{x}=u+v\sqrt{x}$ and you are done (it is an integer iff $u$ is an integer and $v=0$), or $a+b\sqrt{x}$ is not a square in $\Bbb{Q}(\sqrt{x})$ and then the whole thing is certainly not an integer. – Servaes Mar 31 '19 at 11:38
@DanielP It becomes more complicated when you sum more terms, but this is precisely where this approach is nontrivial and does work. – Servaes Mar 31 '19 at 11:39
Ah, nevermind, I'm not too familiar with algebraic fields, I'll read your whole answer again and try to understand it even more. – Daniel P Mar 31 '19 at 11:40
I read your answer again, and I have a couple questions: $1$. How to show that equations of the form $19+6\sqrt{2}=(a+b\sqrt{2})^2$ have a solution? You pointed out that you quickly got $(a,b) = (1,3)$ as the answer. How to get this solution though? $2$. Similarly, "$93+63\sqrt{85}=(a+b\sqrt{85})^2$ is quickly checked to have no rational solutions". How? – Daniel P Mar 31 '19 at 11:51
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Expanding the squares yields $$19+6\sqrt{2}=(a^2+2b^2)+2ab\sqrt{2},$$ and so solving $a^2+2b^2=19$ and $2ab=6$ yields the solutions $(a,b)=(1,3)$ and $(a,b)=(-1,-3)$. Similarly $$93+63\sqrt{85}=(a^2+85b^2)+2ab\sqrt{85},$$ leads to solving $a^2+85b^2=93$ and $2ab=63$, which has no solutions. One way to see this is to note that $b=\tfrac{63}{2a}$ and so $$93=a^2+85b^2=a^2+85\left(\tfrac{63}{2a}\right)^2,$$ but then $a$ is a root of $$4a^4-4\times93a^2+85\times63^2=0,$$ which has no rational (or even real) roots. – Servaes Mar 31 '19 at 12:19
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Though I must admit I did not do any of these calculations; in stead I noted that the norm of $93+63\sqrt{85}$ is negative, so it is not a square. – Servaes Mar 31 '19 at 12:26

How to prove that expessions like $\sqrt{93+63\sqrt{85}} - \sqrt{143} \notin \Bbb{Z}$?

The Problem:

The Question:

3 Answers3