Demonstrate that $\sqrt{\frac{(a-1)(a+3)}{12}}$ is rational, where $a=[(2+\sqrt{3})^n]$.

Question

Demonstrate that $\sqrt{\frac{(a-1)(a+3)}{12}}$ is rational, where $a=[(2+\sqrt{3})^n]$ ($[a]$ represents the floor function of $a$).

My idea

So for the radical to be rational, $\frac{(a-1)(a+3)}{12}$ should have the form $k^2\Rightarrow\frac{(a-1)(a+3)}{3}$ should have the form $k^2\Rightarrow(a-1)(a+3)$ is a multiple of 3+1 and should have the form $3p^2$.

I don't know what else to do forward. Hope one of you can help me!

@AnneBauval Because $\frac{(a-1)(a+3)}{12}$ should have the form $k^2$ and $\frac{1}{4}$ is already a perfect square $(\frac{1}{2})^2$ — IONELA BUCIU, Jan 08 '25 at 10:31
You can try multiplying the numerator and denominator of the fraction inside that square root by $3$ to get: $\sqrt{\frac{3(a - 1)(a + 3)}{36}} = \frac{\sqrt{3(a - 1)(a + 3)}}{6}$, and now the problem is proving the square root inside that numerator is integer. — ducbadatcs, Jan 08 '25 at 10:34
@ducbadatcs yeah this is what i got to too, but how to demonstrate it — IONELA BUCIU, Jan 08 '25 at 10:44
Hint: prove $a=t^n+t^{-n}-1,,\sqrt{\frac{(a-1)(a+3)}{12}}=\frac{t^n-t^{-n}}{2\sqrt{3}}$ with $t^{\pm1}=2\pm\sqrt{3}$. — J.G., Jan 08 '25 at 11:46

score 1 · Answer 1 · edited Jan 08 '25 at 12:49

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Writing $a$ as $2b-1$ shows that you are interested in solutions of $$b^2-1=3x^2$$ The integer solutions of this Pell's Equation are given by $2, 7, 26, ...$.

These satisfy the recurrence relation

$$b_{n+2}-4b_{n+1}+b_n=0$$

which has solution $$2b_n=(2+\sqrt3)^n+(2-\sqrt3)^n$$

Therefore $a=(2+\sqrt3)^n+(2-\sqrt3)^n-1$ will give a rational square root and this is $[(2+\sqrt{3})^n]$, as you require.

edited Jan 08 '25 at 12:49

Gary

36,640

answered Jan 08 '25 at 12:45

user1172706

2,791

@Gary Thanks for the edit. What had I done wrong to make the equation display incorrectly? – user1172706 Jan 08 '25 at 12:50
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You closed a curly bracket with a parenthesis. – Gary Jan 08 '25 at 12:52
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My eyesight must be going. Thanks again. – user1172706 Jan 08 '25 at 12:53
(a-1)(a+3)=(a+1)^2-4 and so this substitution enables us to get the problem into a form where the theory of Pellian Equations can be used. Your final "Why" concerns simply finding a=2b-1 by subtracting 1 from 2b. – user1172706 Jan 08 '25 at 13:25

score 0 · Answer 2 · answered Jan 08 '25 at 14:16

We shall use the fact that $$a=r_+^n+r_-^n-1\quad\text{where}\quad r_\pm=2\pm\sqrt3.$$ By the binomial formula, the two following numbers are integers $$u:=\frac{r_+^n+r_-^n}2,\quad v:=\frac{r_+^n-r_-^n}{2\sqrt3}.$$ From $$1=r_+r_-=r_+^nr_-^n=(u+v\sqrt3)(u-v\sqrt3)=u^2-3v^2=\frac{(a+1)^2}4-3v^2$$ we derive: $$\frac{(a-1)(a+3)}{12}=v^2.$$

Demonstrate that $\sqrt{\frac{(a-1)(a+3)}{12}}$ is rational, where $a=[(2+\sqrt{3})^n]$.

2 Answers2