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

Helen Lewis 2022-01-05 Answered
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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Expert Answer

MoxboasteBots5h
Answered 2022-01-06 Author has 3742 answers
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.
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Mary Goodson
Answered 2022-01-07 Author has 1678 answers
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
\(\displaystyle{\left({1}+\sqrt{{2}}\right)}^{{n}}={a}_{{n}}+{b}_{{n}}\sqrt{{2}}=\sqrt{{{a}^{{{\frac{{{2}}}{{{n}}}}}}}}+\sqrt{{{a}^{{{\frac{{{2}}}{{{n}}}}}}-{1}}}\)
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Vasquez
Answered 2022-01-11 Author has 8850 answers

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

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