Helen Lewis

2022-01-05

Expansion of $(1+\sqrt{2})}^{n$

I was asked to show that$\mathrm{\forall}n\in \mathbb{N}$ there exist a $p\in {\mathbb{N}}^{\cdot}$ 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.

I was asked to show that

I used induction but it wasn't fruitful, so I tried to use the binomial expansion of

Any hint is welcomed.

MoxboasteBots5h

Beginner2022-01-06Added 35 answers

The binomial formula shows you that

$(1+\sqrt{2})}^{n}={a}_{n}+{b}_{n}\sqrt{2$

for some integers${a}_{n},{b}_{n}.$

But, the same binomial formula shows you that (convince yourself of this)

$(1-\sqrt{2})}^{n}={a}_{n}-{b}_{n}\sqrt{2$

for the same integers${a}_{n},{b}_{n}.$

Then comes the hint: Calculate both

$({a}_{n}+{b}_{n}\sqrt{2})({a}_{n}-{b}_{n}\sqrt{2})$

and

$(1+\sqrt{2})}^{n}{(1-\sqrt{2})}^{n}={\left[(1+\sqrt{2})(1-\sqrt{2})\right]}^{n$

and compare.

You will get that$a\frac{2}{n}-2b\frac{2}{n}={(-1)}^{n}$ , 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.

for some integers

But, the same binomial formula shows you that (convince yourself of this)

for the same integers

Then comes the hint: Calculate both

and

and compare.

You will get that

Mary Goodson

Beginner2022-01-07Added 37 answers

This is a bit more roundabout than Jyrkis

Vasquez

Expert2022-01-11Added 669 answers

It's reduced to show that the following expression––––––––––––––––––––– is an integer:

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