What is the pattern in the powers of $\sqrt{2}-\sqrt{1}$?

Question

What is the pattern in this?

$$\begin{align} \left(\sqrt{2}-\sqrt{1}\right)^1 &= \sqrt{2}-\sqrt{1}\\ \left(\sqrt{2}-\sqrt{1}\right)^2 &= \sqrt{9}-\sqrt{8}\\ \left(\sqrt{2}-\sqrt{1}\right)^3 &= \sqrt{50}-\sqrt{49}\\ \left(\sqrt{2}-\sqrt{1}\right)^4 &= \sqrt{289}-\sqrt{288}\\ \end{align}$$

I thought of applying the binomial theorem

Answer 1

Given that

$$\left(\sqrt{2}-\sqrt{1}\right)^k = \sqrt{A_k}-\sqrt{A_k-1}$$

then

$$\left(\sqrt{2}-\sqrt{1}\right)^{k+1} = \left(\sqrt{2}-\sqrt{1}\right)(\sqrt{A_k}-\sqrt{A_k-1})=\\=\sqrt 2\sqrt{A_k}-\sqrt 2\sqrt{A_k-1}-\sqrt{A_k}+\sqrt{A_k-1}=$$

$$=\sqrt{A_{k+1}}-\sqrt{A_{k+1}-1}$$

with

• $\sqrt{A_{k+1}}=\sqrt 2\sqrt{A_k}+\sqrt{A_k-1} \implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$

• $\sqrt{A_{k+1}-1}=\sqrt 2\sqrt{A_k-1}+\sqrt{A_k}\implies A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$

that is

$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)}$$

and $A_1=2$.

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Using an approach similar to this one, we have

$$A_{k+1}=3A_k-1+2\sqrt2\sqrt{A_k(A_k-1)} \\\iff 2A_{k+1}-1=3(2A_{k}-1)+2\sqrt2\sqrt{(2A_k-1)^2-1}$$

and by $2A_{k+1}-1=\frac12\left(t_k+\frac1{t_k}\right) \implies t_1=3+2\sqrt 2$ we obtain

$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\sqrt{\left(t_k+\frac1{t_k}\right)^2-4}$$

$$t_{k+1}+\frac1{t_{k+1}}=3\left(t_k+\frac1{t_k}\right)+2\sqrt 2\left(t_k-\frac1{t_k}\right)$$

$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{3-2\sqrt 2}{t_k}$$

$$t_{k+1}+\frac1{t_{k+1}}=(3+2\sqrt 2)t_k+\frac{1}{(3+2\sqrt 2)t_k}$$

$$t_{k}=(3+2\sqrt 2)^k$$

and then

$$A_k=\frac{(3+2\sqrt 2)^k+(3-2\sqrt 2)^k+2}{4}$$

Answer 2

If $(\sqrt 2 - 1)^n = \sqrt {a_n } - \sqrt {a_n - 1}$, then $ (\sqrt 2 + 1)^n = \sqrt {a_n } + \sqrt {a_n - 1}$. Hence, $$ a_n = \left( {\frac{{(\sqrt 2 + 1)^n + (\sqrt 2 - 1)^n }}{2}} \right)^2 = \frac{{2 + (3 + 2\sqrt 2 )^n + (3 - 2\sqrt 2 )^n }}{4}. $$ This is A115599 in the OEIS. See also A055997. You can show using the binomial theorem that $a_n$ is always an integer. In fact $$ a_n = \frac{{3^n + 1}}{2} + \sum\limits_{k = 1}^{\left\lfloor {n/2} \right\rfloor } \binom{n}{2k}2^{3k - 1} 3^{n - 2k} . $$

Answer 3

We have $$(\sqrt{1}-\sqrt{2})^k = \sqrt{a_k^2} - \sqrt{2b_k^2}$$ where $a_k,b_k$ obey the following recurrence relationships: $$a_{k+1}=a_{k}+2b_{k}\\ b_{k+1}=a_{k}+b_{k}$$ with $a_{1}=1$ and $b_1 = 1$. This can be proved inductively. To understand the recurrence relationship we can recast it as a matrix equation: $${\begin{pmatrix} a_{k+1}\\ b_{k+1} \end{pmatrix}} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} a_k\\ b_k \end{pmatrix} $$

The eigenvectors are $\begin{pmatrix} \sqrt{2} \\ 1 \end{pmatrix}$, $\begin{pmatrix} \sqrt{2} \\ -1 \end{pmatrix}$ with coresponding eigenvalues: $1+\sqrt{2}$ and $1-\sqrt{2}$.
This allows us to solve: $${\begin{pmatrix} a_{n}\\ b_{n} \end{pmatrix}} = {\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}}^{n-1} \begin{pmatrix} 1\\ 1 \end{pmatrix} $$ with: $$a_{n} = \frac{(1+\sqrt{2})^n + (1-\sqrt{2})^n}{2}\\ b_{n} = \frac{(1+\sqrt{2})^n - (1-\sqrt{2})^n}{2\sqrt{2}}$$

A recurrence relationship can also be written for each term separately as:

$$a_{n} = 2a_{n-1} + a_{n-2}\\ b_{n} = 2 b_{n-1} + b_{n-2}$$

with starting terms $a_1 = 1$, $a_2 = 3$, $b_1 = 1$, $b_2 = 2$.

We can write it as a recurrence relationship or a matrix equation if we wish to use only integers, or we can write the terms individually, but that causes the reappearance of $\sqrt{2}$ in the formula.