Proving $\sqrt{4+2\sqrt{3}} - \sqrt{3}$ is rational

Question

Problem: Show that $\sqrt{4+2\sqrt{3}} - \sqrt{3} \in \mathbb{Q}$

My approach

By definition, $\sqrt{4+2\sqrt{3}} - \sqrt{3} \in \mathbb{Q} \iff \exists n,m \in \mathbb{Z}: \sqrt{4+2\sqrt{3}} - \sqrt{3} = \frac{n}{m}$

Carrying out the algebra, we get:

$m\sqrt{4+2\sqrt{3}} = n+m\sqrt{3} \implies m^2(4+2\sqrt{3}) = n^2 + 2nm\sqrt{3} + 3m^2$

$\implies m^2 + (2m^2-2mn)\sqrt{3} - n^2 = 0$

We need to find integer solutions for $m,n$ to show that this expression evaluates to a rational number. In this case, we can choose $m=n$ for any $n \in \mathbb{Z}$, so we have the equivalency class $\{1\}$ as the set of values for $m,n$ that solve this equation..

Is this a valid proof or am I making an implicit assumptions or not checking a condition?

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Edit on completing the above proof

So I conferred with a mathematics professor, and he confirmed my reasoning is valid, but incomplete. In my above proof, I square the expressions in one of my steps, which introduces extraneous solutions (i.e., my steps are not reversible).

Therefore, all I have shown is that $1$ may be the value of $\sqrt{4+2\sqrt{3}} - \sqrt{3}$. The proof now requires me to show that, indeed, $\sqrt{4+2\sqrt{3}} - \sqrt{3} =1$.

Thankfully, this is an easy final check:

$$\sqrt{4+2\sqrt{3}} - \sqrt{3} = 1 \implies \sqrt{4+2\sqrt{3}} =1+ \sqrt{3} \implies 4+2\sqrt{3} = 1+2\sqrt{3} + 3 = 4+2\sqrt{3}$$

$\square$

Answer 1

$$\left( ~1 + \sqrt{3} ~\right)^2 = 4 + 2\sqrt{3} \implies $$

$$\sqrt{4 + 2\sqrt{3}} = 1 + \sqrt{3}.$$

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The above analysis has the defect of being an answer rather than a (derived) solution.

Suppose

$$\left( a + b\sqrt{3} ~\right)^2 = 4 + 2\sqrt{3}.$$

Then $~2ab = 2~$ and $~a^2 + 3b^2 = 4.$

This implies that

$$b = \frac{1}{a} \implies a^2 + \frac{3}{a^2} = 4.$$

This implies that

$$a^4 - 4a^2 + 3 = 0 \implies (a^2 - 3) \times (a^2 - 1) = 0.$$

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$\underline{\text{Addendum}}$

$$\implies m^2 + (2m^2-2mn)\sqrt{3} - n^2 = 0$$

If $~r~$ and $~s~$ are integers and

$$r + s\sqrt{3} = 0,$$

then $~r = 0~$ and $~s = 0.$

So, the original poster's approach, up to the excerpted line above is viable.

You end up with $~m = n~$ which implies that

$$\sqrt{4 + \sqrt{3}} - \sqrt{3} = 1, \tag1 $$

from which the problem can be completed by an alternative route.

As a clarification, (1) above does not prove that $~m = n.~$ Instead, it proves that if $~\displaystyle \sqrt{4 + \sqrt{3}} - \sqrt{3}~$ is rational, then it must be equal to $~1.$

Answer 2

Suppose $$\sqrt{4+2\sqrt{3}} - \sqrt{3}=r \notin \mathbb{Q}\;\;\;(*)$$

Particulary, this means $r\ne 1$. Then $$4+2\sqrt{3} = (r+\sqrt{3})^2$$

so $$4+2\sqrt{3} = 3+2r\sqrt{3}+r^2$$

so $$2\sqrt{3}(1-r) = r^2-1$$

Since $r$ can not be $1$ we have $$r= -2\sqrt{3}-1$$

But then if we plugg this in $(*)$ we get

$$ \sqrt{4+2\sqrt{3}} = -\sqrt{3}-1<0$$

A contradiction.

Answer 3

This is exactly $1$: $$ \begin{align} \sqrt{4+2\sqrt3}=&\sqrt{1^2+\sqrt3^2+2\times1\times\sqrt3}\\ =&\sqrt{(1+\sqrt3)^2}\\ =&1+\sqrt3 \end{align} $$

Answer 4

Because m and n are integers, you can find that $m = n$ by equalizing all the coefficients. $$m^2(4+2\sqrt{3}) = n^2 + 2nm\sqrt{3} + 3m^2$$ $$4m^2+2m^2\sqrt{3} = 3m^2 + n^2 + 2mn\sqrt{3}$$ Then: $$\left\{\begin{matrix} 4m^2=3m^2+n^2 \\ 2m^2=2mn \end{matrix}\right.$$ implying that $m = n$.

Bonus: You can also do the same thing starting from the equation that you get. Because the RHS is 0, all the coefficients from the LHS should be 0 as well.