# Expansion of (1+\sqrt2)^n I was asked to show that

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

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MoxboasteBots5h
The binomial formula shows you that
$$\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}={a}_{{n}}+{b}_{{n}}\sqrt{{2}}$$
for some integers $$\displaystyle{a}_{{n}},{b}_{{n}}.$$
But, the same binomial formula shows you that (convince yourself of this)
$$\displaystyle{\left({1}-\sqrt{{2}}\right)}^{{n}}={a}_{{n}}-{b}_{{n}}\sqrt{{2}}$$
for the same integers $$\displaystyle{a}_{{n}},{b}_{{n}}.$$
Then comes the hint: Calculate both
$$\displaystyle{\left({a}_{{n}}+{b}_{{n}}\sqrt{{2}}\right)}{\left({a}_{{n}}-{b}_{{n}}\sqrt{{2}}\right)}$$
and
$$\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}{\left({1}-\sqrt{{2}}\right)}^{{n}}={\left[{\left({1}+\sqrt{{2}}\right)}{\left({1}-\sqrt{{2}}\right)}\right]}^{{n}}$$
and compare.
You will get that $$\displaystyle{a}{\frac{{{2}}}{{{n}}}}-{2}{b}{\frac{{{2}}}{{{n}}}}={\left(-{1}\right)}^{{n}}$$, so $$\displaystyle{a}{\frac{{{2}}}{{{n}}}}$$ and $$\displaystyle{2}{b}{\frac{{{2}}}{{{n}}}}$$ differ from each other by one, and $$\displaystyle{p}$$ will be the larger of the two.
###### Not exactly what youâ€™re looking for?
Mary Goodson
This is a bit more roundabout than Jyrki's answer, but I have a soft spot for linear recurrences.
First, prove that $$\displaystyle{\left({1}+\sqrt{{{2}}}\right)}^{{n}}={a}_{{n}}+{b}^{{n}}\sqrt{{2}}$$ where
$$\displaystyle{a}_{{0}}={1},{a}_{{1}}={1},{a}_{{n}}={2}{a}_{{{n}-{1}}}+{a}_{{{n}-{2}}}$$
$$\displaystyle{b}_{{0}}={0},{b}_{{1}}={1},{b}_{{n}}={2}{b}_{{{n}-{1}}}+{b}_{{{n}-{2}}}$$
Then prove that
$$\displaystyle{a}_{{n}}={\frac{{{1}}}{{{2}}}}{\left({1}+\sqrt{{2}}\right)}^{{n}}+{\frac{{{1}}}{{{2}}}}{\left({1}-\sqrt{{2}}\right)}^{{n}}$$
$$\displaystyle{b}_{{n}}={\frac{{{1}}}{{{2}\sqrt{{2}}}}}{\left({1}+\sqrt{{2}}\right)}^{{n}}-{\frac{{{1}}}{{{2}\sqrt{{2}}}}}{\left({1}-\sqrt{{2}}\right)}^{{n}}$$
Now conclude that $$\displaystyle{a}^{{{\frac{{{2}}}{{{n}}}}}}={2}{b}^{{{\frac{{{2}}}{{{n}}}}}}+{1}$$, so that
$$\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}={a}_{{n}}+{b}_{{n}}\sqrt{{2}}=\sqrt{{{a}^{{{\frac{{{2}}}{{{n}}}}}}}}+\sqrt{{{a}^{{{\frac{{{2}}}{{{n}}}}}}-{1}}}$$
Vasquez

It's reduced to show that the following expression––––––––––––––––––––– is an integer:
$$[\frac{(1+\sqrt2)^{-n}}{2}]^2=\frac{1}{4}[(1+\sqrt2)^{2n}+(1-\sqrt2)^{2n}]+\frac{1}{2} \\\frac{1}{2}+\frac{1}{4}[(1+\sqrt{2})^{2n}+(1-\sqrt2)^{2n}] \\=\frac{1}{2}+\frac{1}{4}[\sum_{k=0}^{2n}((2n),(k))2^{\frac{k}{2}}+\sum_{k=0}^{2n}((2n),(k))(-1)^k2^{\frac{k}{2}}] \\=\frac{1}{2}+\frac{1}{4}[2\sum_{k=0}^{n}((2n),(2k))2^k]=\frac{1}{2}+\frac{1}{2}[1+\sum_{k=1}^n((2n),(2k))2^k] \\=1+\sum_{k=1}^n((2n),(2k))2^{k-1}$$