jmc

Let $\alpha =2+\sqrt{3}$ and $\beta =2-\sqrt{3}.$ Then consider the number $$\begin{array}{rl}N& ={\alpha }^{1000}+{\beta }^{1000}\\ & =(2+\sqrt{3}{)}^{1000}+(2-\sqrt{3}{)}^{1000}\\ & =2^{1000}+\left(\genfrac{}{}{0ex}{}{1000}{1}\right)2^{999}(\sqrt{3})+\left(\genfrac{}{}{0ex}{}{1000}{2}\right)2^{998}(\sqrt{3}{)}^2+(\genfrac{}{}{0ex}{}{1000}{3})(\sqrt{3}{)}^3+\cdots \\ & +2^{1000}-\left(\genfrac{}{}{0ex}{}{1000}{1}\right)2^{999}(\sqrt{3})+\left(\genfrac{}{}{0ex}{}{1000}{2}\right)2^{998}(\sqrt{3}{)}^2-(\genfrac{}{}{0ex}{}{1000}{3})(\sqrt{3}{)}^3+\cdots .\end{array}$$Adding $(2+\sqrt{3}{)}^{1000}$ and $(2-\sqrt{3}{)}^{1000}$, we see that all the terms containing a $\sqrt{3}$ will cancel, meaning that we are left with an integer.

Furthermore, $$\beta =2-\sqrt{3}=\frac{(2-\sqrt{3})(2+\sqrt{3})}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}<1,$$so $0<{\beta }^{1000}<1.$

Therefore, $$N-1<{\alpha }^{1000}<N,$$which means $n=\lfloor {\alpha }^{1000}\rfloor =N-1.$

Then $$f=x-n={\alpha }^{1000}-(N-1)=1-{\beta }^{1000},$$so $1-f={\beta }^{1000}.$ Hence, $$\begin{array}{rl}x(1-f)& ={\alpha }^{1000}{\beta }^{1000}\\ & =(\alpha \beta {)}^{1000}\\ & =[(2+\sqrt{3})(2-\sqrt{3}){]}^{1000}\\ & =1^{1000}\\ & =\boxed{1}.\end{array}$$