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Let $n$ be a positive integer. Show that $(\sqrt{2}+1)^n = \sqrt{m}+\sqrt{m-1}$ for some positive integer $m$.

I was wondering what would be the easiest way of proving this. Expanding $(\sqrt{2}+1)^n$ we get $$(\sqrt{2}+1)^n = \binom{n}{0}(\sqrt{2})^n+\binom{n}{1}(\sqrt{2})^{n-1}+\cdots+\binom{n}{n} = a+b\sqrt{2}$$ and so either $\sqrt{m} = a$ and $\sqrt{m-1} = b\sqrt{2}$ or $\sqrt{m} = b\sqrt{2}$ and $\sqrt{m-1} = a$. How do we continue from here?

Math1000
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user19405892
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    I haven't checked if it is true. How about trying induction? – Sayan Oct 15 '16 at 11:30
  • For a hint: suppose $a^2=2b^2\pm 1$, then $(a+2b)^2=a^2+4ab+4b^2=2a^2+4ab+2b^2\mp 1=2(a+b)^2\mp 1$. You just need to work with the various expressions with confidence and you should be able to make an induction work. – Mark Bennet Oct 15 '16 at 11:39
  • Hint: $(1+\sqrt{2})^n = a + b\sqrt{2} \implies (1-\sqrt{2})^n = a - b\sqrt{2} \implies a^2 - 2b^2 = (-1)^n$. – achille hui Oct 15 '16 at 12:09

1 Answers1

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You can try to prove this by induction:

For every natural $n$, $(1+\sqrt 2)^n=\sqrt{m}+\sqrt{m-1}$ where:

  • If $n$ is odd, $m/2$ and $m-1$ are perfect squares
  • If $n$ is even, $m$ and $(m-1)/2$ are perfect squares

This can seem more difficult, because you have to prove more things, but you also have more information for the inductive step.

ajotatxe
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