Helen Lewis

2022-01-05

Expansion of ${\left(1+\sqrt{2}\right)}^{n}$
I was asked to show that $\mathrm{\forall }n\in \mathbb{N}$ there exist a $p\in {\mathbb{N}}^{\cdot }$ such that
${\left(1+\sqrt{2}\right)}^{n}=\sqrt{p}+\sqrt{p-1}$
I used induction but it wasn't fruitful, so I tried to use the binomial expansion of ${\left(1+\sqrt{2}\right)}^{n}$ but it seems I lack some insight to go further.
Any hint is welcomed.

MoxboasteBots5h

The binomial formula shows you that
${\left(1+\sqrt{2}\right)}^{n}={a}_{n}+{b}_{n}\sqrt{2}$
for some integers ${a}_{n},{b}_{n}.$
But, the same binomial formula shows you that (convince yourself of this)
${\left(1-\sqrt{2}\right)}^{n}={a}_{n}-{b}_{n}\sqrt{2}$
for the same integers ${a}_{n},{b}_{n}.$
Then comes the hint: Calculate both
$\left({a}_{n}+{b}_{n}\sqrt{2}\right)\left({a}_{n}-{b}_{n}\sqrt{2}\right)$
and
${\left(1+\sqrt{2}\right)}^{n}{\left(1-\sqrt{2}\right)}^{n}={\left[\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)\right]}^{n}$
and compare.
You will get that $a\frac{2}{n}-2b\frac{2}{n}={\left(-1\right)}^{n}$, so $a\frac{2}{n}$ and $2b\frac{2}{n}$ differ from each other by one, and $p$ will be the larger of the two.

Mary Goodson

This is a bit more roundabout than Jyrkis

Vasquez

It's reduced to show that the following expression––––––––––––––––––––– is an integer:
$\left[\frac{\left(1+\sqrt{2}{\right)}^{-n}}{2}{\right]}^{2}=\frac{1}{4}\left[\left(1+\sqrt{2}{\right)}^{2n}+\left(1-\sqrt{2}{\right)}^{2n}\right]+\frac{1}{2}\phantom{\rule{0ex}{0ex}}\frac{1}{2}+\frac{1}{4}\left[\left(1+\sqrt{2}{\right)}^{2n}+\left(1-\sqrt{2}{\right)}^{2n}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}+\frac{1}{4}\left[\sum _{k=0}^{2n}\left(\left(2n\right),\left(k\right)\right){2}^{\frac{k}{2}}+\sum _{k=0}^{2n}\left(\left(2n\right),\left(k\right)\right)\left(-1{\right)}^{k}{2}^{\frac{k}{2}}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{2}+\frac{1}{4}\left[2\sum _{k=0}^{n}\left(\left(2n\right),\left(2k\right)\right){2}^{k}\right]=\frac{1}{2}+\frac{1}{2}\left[1+\sum _{k=1}^{n}\left(\left(2n\right),\left(2k\right)\right){2}^{k}\right]\phantom{\rule{0ex}{0ex}}=1+\sum _{k=1}^{n}\left(\left(2n\right),\left(2k\right)\right){2}^{k-1}$

Do you have a similar question?