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If $$(7+4\sqrt{3})^n = p+\beta,$$ where $n$ and $p$ are positive integers and $\beta$ is a proper fraction, then show that $$(1-\beta)(p+\beta)=1.$$

I cant even understand how to express the term in a positive number and a proper fraction. I would appreciate any hint.

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  • $\begingroup$ By "proper fraction", do we mean a real number between $0$ and $1$? $\endgroup$ – 6005 Nov 7 '16 at 3:03
  • $\begingroup$ Yeah, the proper fraction means that $\endgroup$ – user354545 Nov 7 '16 at 3:08
  • $\begingroup$ Hint: $(7+4\sqrt{3})(7-4\sqrt{3}) = 1$. $\endgroup$ – dxiv Nov 7 '16 at 3:20
  • $\begingroup$ No its $1-beta$ $\endgroup$ – user354545 Nov 7 '16 at 3:21
  • $\begingroup$ @dxiv No, it is $1 - \beta$. $\endgroup$ – 6005 Nov 7 '16 at 3:22
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Fun question! The key realization is that $$ (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n \in \mathbb{Z} $$ (do you see why?) and moreover, that $$ 0 < (7 - 4\sqrt{3}) < 1, $$ so that $$ 0 < (7 - 4\sqrt{3})^n < 1, $$ for all natural numbers $n$. It follows from here that \begin{align*} p &= (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n - 1 \\ \beta &= 1 - (7 - 4\sqrt{3})^n. \end{align*} Now to finish, we see directly from the above that \begin{align*} p + \beta &= (7 + 4\sqrt{3})^n \\ 1 - \beta &= (7 - 4\sqrt{3})^n. \end{align*} Multiply them together and see what you get.

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  • $\begingroup$ I didnt get how did we get values of p and beta $\endgroup$ – user354545 Nov 7 '16 at 3:29
  • $\begingroup$ I took p as (7+4√3)^n + (7-4√3)^n and beta as (7-4√3)^n which becomes (7+4√3)^n when we add them. But when i solve it, i dont get the answer $\endgroup$ – user354545 Nov 7 '16 at 3:34
  • $\begingroup$ Oh , i get it we actually added 1 in the p term and subtracted 1 from the beta term, their meanings remain the same still $\endgroup$ – user354545 Nov 7 '16 at 3:37
  • $\begingroup$ @user354545 Yes. You should check that the $\beta$ I gave is between $0$ and $1$, and that $p + \beta$ is $(7 + 4\sqrt{3})^n$. That alone means that we picked the right $p$ and $\beta$ $\endgroup$ – 6005 Nov 7 '16 at 4:18
  • $\begingroup$ Yeah but beta = 1-(7+4√3)^n acheives the value 1 when n is a higher power $\endgroup$ – user354545 Nov 7 '16 at 4:20

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