I was asked to show that $\forall n\in \mathbb N$ there exist a $p\in \mathbb N^\ast$ such that $$(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1}$$
I used induction but it wasn't fruitful, so I tried to use the binomial expansion of $(1+\sqrt{2})^n$ but it seems I lack some insight to go further.
Any hint is welcomed.
The binomial formula shows you that $$(1+\sqrt2)^n=a_n+b_n\sqrt2$$ for some integers $a_n, b_n$.
But, the same binomial formula shows you that (convince yourself of this) $$ (1-\sqrt2)^n=a_n-b_n\sqrt2 $$ for the same integers $a_n,b_n$.
Then comes the hint: Calculate both $$(a_n+b_n\sqrt2)(a_n-b_n\sqrt2)$$ and $$(1+\sqrt2)^n(1-\sqrt2)^n=[(1+\sqrt2)(1-\sqrt2)]^n$$ and compare.
You will get that $a_n^2-2b_n^2=(-1)^n$, so $a_n^2$ and $2b_n^2$ differ from each other by one, and $p$ will be the larger of the two.
This is a bit more roundabout than Jyrki's answer, but I have a soft spot for linear recurrences.
First, prove that $(1+\sqrt{2})^n=a_n+b_n\sqrt{2}$ where $$a_0=1,\quad a_1=1,\quad a_n=2a_{n-1}+a_{n-2}\\ b_0=0,\quad b_1=1,\quad b_n=2b_{n-1}+b_{n-2}$$ Then prove that $$a_n=\tfrac{1}{2} (1 + \sqrt{2})^n + \tfrac{1}{2}(1 - \sqrt{2})^n\\ b_n=\tfrac{1}{2\sqrt{2}} (1 + \sqrt{2})^n - \tfrac{1}{2\sqrt{2}}(1 - \sqrt{2})^n$$ Now conclude that $a_n^2=2b_n^2+1$, so that $$(1+\sqrt{2})^n=a_n+b_n\sqrt{2}=\sqrt{a_n^2}+\sqrt{a_n^2-1}$$
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ It's reduced to show that the $\ul{following\ expression}$ is an integer: \begin{align} \bracks{\pars{1 + \root{2}}^{n} + \pars{1 + \root{2}}^{-n} \over 2}^{2} = {1 \over 4}\bracks{\pars{1 + \root{2}}^{2n} + \pars{1 - \root{2}}^{2n}} + \half \end{align}
Indeed, the right hand side is $\ds{\ul{the\ value}\ \mbox{of}\ p}$: $$ \color{#f00}{\pars{1 + \root{2}}^{n}} = \color{#f00}{\root{1 + \sum_{k = 1}^{n}{2n \choose 2k}2^{k - 1}} + \root{\sum_{k = 1}^{n}{2n \choose 2k}2^{k - 1}}} $$
For any $n\in\mathbb{N}$, $(1+\sqrt{2})^n$ is an algebraic number over $\mathbb{Q}$ with degree $\leq 2$, so, assuming that equality holds, $\sqrt{p}+\sqrt{p-1}$ has to be an algebraic number over $\mathbb{Q}$ with degree $\leq 2$. No issues with $p=1$ and $p=2$, but if $p\geq 3$ and neither $p$ or $p-1$ is a square... let us see. Since
$$ (\sqrt{p}+\sqrt{p-1})^2 = 2p-1+2\sqrt{p(p-1)} $$
the biquadratic polynomial
$$ q(x) = x^4+(2-4p)x^2+1 $$
vanishes at $x=\sqrt{p}+\sqrt{p-1}$. If we prove that $q(x)$ is the minimal polynomial of $\sqrt{p}+\sqrt{p-1}$ over $\mathbb{Q}$ we are done and doomed, since we have that $\sqrt{p}+\sqrt{p-1}$
is an algebraic number of degree $4$ over $\mathbb{Q}$.
The roots of $q(x)$ are given by $\pm\sqrt{p}\pm\sqrt{p-1}$, but for every choice of signs $\varepsilon_i$
$$ (x-\varepsilon_1\sqrt{p}-\varepsilon_2\sqrt{p-1})(x-\varepsilon_3\sqrt{p}-\varepsilon_4\sqrt{p-1})\not\in\mathbb{Q}[x]$$
by Vieta's formulas, so $q(x)$ is actually the minimal polynomial of $\sqrt{p}+\sqrt{p-1}$ over $\mathbb{Q}$ and the original problem boils down to checking if equality is achieved by some $n$ just in the cases for which $p$ is a square or a square plus one. However, it is easy to check that by expanding
$$ (1+\sqrt{2})^n = a_n+b_n\sqrt{2} $$
we must have $a_n^2=p$ and $2b_n^2=p-1$ or $a_n^2=p-1$ and $2b_n^2=p$. The ratio $\frac{a_n}{b_n}$ of the Lucas-Pell numbers $a_n,b_n$ converges pretty fast to $\sqrt{2}$, and since $\color{red}{a_n^2-2b_n^2=(-1)^n}$, we have:
$$ (1+\sqrt{2})^{2n} = \sqrt{a_{2n}^2} + \sqrt{a_{2n}^2-1} $$
and
$$ (1+\sqrt{2})^{2n+1} = \sqrt{2b_{2n+1}^2}+\sqrt{2b_{2n+1}^2-1}.$$
Note: I added some lines to show this explicit formula: $p =1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1} $.
If $(1+\sqrt{2})^n = \sqrt{p} + \sqrt{p-1} $, then $(1+\sqrt{2})^{2n} = 2p-1+2\sqrt{p(p-1)} $ so that
$\begin{array}\\ 4p(p-1) &=((1+\sqrt{2})^{2n} - (2p-1))^2\\ &=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)+(2p-1)^2\\ &=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)+4p^2-4p+1\\ \text{so that}\\ -1 &=(1+\sqrt{2})^{4n}-2(1+\sqrt{2})^{2n} (2p-1)\\ \implies\\ 2p-1 &=\frac{(1+\sqrt{2})^{4n}+1}{2(1+\sqrt{2})^{2n} }\\ &=\frac12(1+\sqrt{2})^{2n}+\frac{1}{2(1+\sqrt{2})^{2n} }\\ &=\frac12(1+\sqrt{2})^{2n}+\frac12(-1+\sqrt{2})^{2n} \qquad\text{ since} (1+\sqrt{2})(-1+\sqrt{2})=1\\ &=\frac12\sum_{k=0}^{2n} \binom{2n}{k}(\sqrt{2})^{2n-k}(1+(-1)^k)\\ &=\sum_{k=0}^{n} \binom{2n}{2k}(\sqrt{2})^{2n-2k}\\ &=\sum_{k=0}^{n} \binom{2n}{2k}2^{n-k}\\ &=\binom{2n}{2n}2^{n-n}+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k}\\ &=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k}\\ &=1+2\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}\\ \text{so}\\ p &=1+\sum_{k=0}^{n-1} \binom{2n}{2k}2^{n-k-1}\\ \end{array} $
$$6a_n=[(3-2\sqrt{2})^n+(3+2\sqrt{2})^n][(3-2\sqrt{2})+(3+2\sqrt{2})]$$ $$6a_n=a_{n-1}+a_{n+1}$$ Converting to $p_n$ as $4p_n-2=a_n$, we have $6p_{n}=p_{n+1}+p_{n-1}+2$, with $p_1=2$ and $p_2=9$. Now its easy to see that $p_n$ is always a positive integer :)
– Sawarnik Jul 26 '16 at 20:09