Show that $(1-\beta)(p+\beta)=1$

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.

• By "proper fraction", do we mean a real number between $0$ and $1$? – 6005 Nov 7 '16 at 3:03
• Yeah, the proper fraction means that – user354545 Nov 7 '16 at 3:08
• Hint: $(7+4\sqrt{3})(7-4\sqrt{3}) = 1$. – dxiv Nov 7 '16 at 3:20
• No its $1-beta$ – user354545 Nov 7 '16 at 3:21
• @dxiv No, it is $1 - \beta$. – 6005 Nov 7 '16 at 3:22

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.
• @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 $\beta$ – 6005 Nov 7 '16 at 4:18