Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem

What I've done so far:

Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$

Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$

and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$

I did this because in a similar example in class, we related $r^2$ and $r^4$ to find a polynomial such that $mr^4+nr^2 = 0$ for some integers $m,n$. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems.

Solution 1:

Let $\displaystyle N=\sqrt{4+2\sqrt{3}}-\sqrt{3}$

$=\sqrt{{(\sqrt{3})}^2+1^2+2\cdot\sqrt{3}\cdot1}-\sqrt{3}$

$=(\sqrt{3}+1-\sqrt{3})$

$=\boxed1$

Aliter: If you want to use polynomials, you can see that

$\displaystyle (N+\sqrt{3})^2=4+2\sqrt{3}$

$\implies N^2+2\sqrt{3}\cdot N + 3=4+2\sqrt{3}$

$\implies (N^2-1)=2\sqrt{3}\cdot(1-N)$

$\implies N=-2\sqrt{3}-1$

or

$N=1$

But since $N>0$,

$\implies N=\boxed1$

Solution 2:

From $r = \sqrt{4+2\sqrt{3}}-\sqrt{3}$ we get $r+\sqrt{3}=\sqrt{4+2\sqrt{3}}$ and, squaring both sides, $$ r^2+2r\sqrt{3}+3=4+2\sqrt{3} $$ and so $$ r^2-1=2(1-r)\sqrt{3} $$ Square again: $$ r^4-2r^2+1=12-24r+12r^2 $$ so $$ r^4-14r^2+24r-11=0 $$ The rational root test only allows $1$, $-1$, $11$ and $-11$ as roots. Since $1$ is clearly a root we have $$ (r-1)(r^3+r^2-13r+11)=0 $$ and $1$ is a root also of the second factor: $$ (r-1)^2(r^2+2r-11)=0 $$ The roots of the second factor are $$ -1+2\sqrt{3},\qquad -1-2\sqrt{3} $$ Since $r>0$, we only have two possibilities: $r=1$ or $r=2\sqrt{3}-1$. The second possibility gives $$ \sqrt{4+2\sqrt{3}}=3\sqrt{3}-1 $$ If we square this, we get $$ 4+2\sqrt{3}=28-6\sqrt{3} $$ or $$ 24=8\sqrt{3} $$ which is absurd. Thus we only remain with the possibility that $r=1$.

Easier: $4+2\sqrt{3}=3+2\sqrt{3}+1=(\sqrt{3}+1)^2$.

Alternatively, from $r^2-1=2(1-r)\sqrt{3}$ we deduce $r=1$ or $$ r+1=-2\sqrt{3} $$ that's absurd, because $r>0$.

Solution 3:

is there another approach to these type of problems

Since $$4=1+3,3=1\times 3$$ we can have $$4+2\sqrt 3=1+3+2\sqrt{1\times 3}=(1+\sqrt 3)^2$$