21
$\begingroup$

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.

$\endgroup$
6
  • 2
    $\begingroup$ What is the $\large *$ meaning ?. $\endgroup$ Jul 24, 2016 at 21:48
  • 3
    $\begingroup$ @FelixMarin people who define $\Bbb N$ as $\{0,1,\ldots\}$ use $\Bbb N^*$ for $\Bbb N \setminus \{0\}$. $\endgroup$
    – user258700
    Jul 24, 2016 at 21:52
  • $\begingroup$ FYI, the values $p=9$ and $n=2$ happen to work $\endgroup$
    – imranfat
    Jul 24, 2016 at 22:29
  • 1
    $\begingroup$ Let $p$ be any real such that $(\sqrt{2}+1)^n = \sqrt{p} + \sqrt{p-1}$. Taking conjugates we have, $(\sqrt{2}-1)^n = \sqrt{p} - \sqrt{p-1}$. Adding them up, we have $\frac12[(\sqrt{2}-1)^n+(\sqrt{2}+1)^n]=\sqrt{p}$. Hence, $$p=\frac14[(3-2\sqrt{2})^n+(3+2\sqrt{2})^n+2]$$. Now you can show that this is an integer, maybe using induction :) $\endgroup$
    – Sawarnik
    Jul 26, 2016 at 19:30
  • 1
    $\begingroup$ Yes, we can do it by induction. Take: $$a_n=(3-2\sqrt{2})^n+(3+2\sqrt{2})^n$$ $$6a_n=[(3-2\sqrt{2})^n+(3+2\sqrt{2})^n][(3-2\sqrt{2})+(3+2\sqrt{2})]$$ $$6a_n=a_{n-1}+a_{n+1}$$ 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 :) $\endgroup$
    – Sawarnik
    Jul 26, 2016 at 20:09

5 Answers 5

36
$\begingroup$

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.

$\endgroup$
5
  • 5
    $\begingroup$ Why are there so many downvotes? This solution is quite correct... $\endgroup$
    – shalop
    Jul 24, 2016 at 21:36
  • 3
    $\begingroup$ @Shalop: I don't know. Looks like somebody disproves of my actions. Either with this question or in general (possibly as a consequence of things I have done as a moderator). I'm not gonna worry about it :-) $\endgroup$ Jul 24, 2016 at 21:38
  • $\begingroup$ @Jyrki Lahtonen simple and so useful (+1) $\endgroup$ Jul 24, 2016 at 23:28
  • $\begingroup$ @JyrkilLahtonen: Nice answer (+1); also as compensation for nonsensical downvotes. $\endgroup$
    – epi163sqrt
    Jul 25, 2016 at 7:03
  • 1
    $\begingroup$ (+1) I posted my answer, then realized yours was essentially the same. $\endgroup$
    – robjohn
    Jul 25, 2016 at 13:43
9
$\begingroup$

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 $$(1+\sqrt{2})^n=a_n+b_n\sqrt{2}=\sqrt{a_n^2}+\sqrt{a_n^2-1}$$

$\endgroup$
1
  • 1
    $\begingroup$ Write $(1+\sqrt 2)^{n+1}=a_{n+1}+b_{n+1}\sqrt 2=(1+\sqrt 2)(a_n+b_n\sqrt 2)=(a_n+2b_n)+(a_n+b_n)\sqrt 2$ $\endgroup$ Jul 24, 2016 at 23:38
7
$\begingroup$

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}


\begin{align} &\color{#f00}{\half + {1 \over 4}\bracks{\pars{1 + \root{2}}^{2n} + \pars{1 - \root{2}}^{2n}}} \\[5mm] & = \half + {1 \over 4}\bracks{\sum_{k = 0}^{2n}{2n \choose k}2^{k/2} + \sum_{k = 0}^{2n}{2n \choose k}\pars{-1}^{k}2^{k/2}} \\[5mm] & = \half + {1 \over 4}\bracks{2\sum_{k = 0}^{n}{2n \choose 2k}2^{k}} = \half + \half\bracks{1 + \sum_{k = 1}^{n}{2n \choose 2k}2^{k}} = \color{#f00}{1 + \sum_{k = 1}^{n}{2n \choose 2k}2^{k - 1}} \end{align} $\ul{which\ is\ an\ integer\,\,\,}$.

Indeed, the right hand side is $\ds{\ul{the\ value}\ \mbox{of}\ p}$: $$ \color{#f00}{\pars{1 + \root{2}}^{n}} = \color{#f00}{\root{1 + \sum_{k = 1}^{n}{2n \choose 2k}2^{k - 1}} + \root{\sum_{k = 1}^{n}{2n \choose 2k}2^{k - 1}}} $$

$\endgroup$
4
  • $\begingroup$ Your answers have lovely complexities. (+1) $\endgroup$ Jul 24, 2016 at 23:36
  • 1
    $\begingroup$ @BehrouzMaleki Thanks. If $a \equiv \left(1 + \sqrt{2}\right)^{n}$, then $p =\left[{1 \over 2}\left(a + {1 \over a}\right)\right]^{2} ={1 \over 4}\left(a^{2} + {1 \over a^{2}}\right) + {1 \over 2}$. I didn't mention this step. $\endgroup$ Jul 24, 2016 at 23:41
  • $\begingroup$ Ah, a constructive proof. Gorgeous. $\endgroup$ Jul 25, 2016 at 4:08
  • $\begingroup$ @BrevanEllefsen Thanks for your remark. $\endgroup$ Jul 25, 2016 at 5:09
6
$\begingroup$

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 $$ (1+\sqrt{2})^{2n+1} = \sqrt{2b_{2n+1}^2}+\sqrt{2b_{2n+1}^2-1}.$$

$\endgroup$
1
  • $\begingroup$ New approach (+1) $\endgroup$ Jul 24, 2016 at 23:28
3
$\begingroup$

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} $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.