Number Theory Problem involving fractional part of a number If 
$x = ( 9 + 4 \sqrt {5} )^{48}$
where $x = [x] + f$, where $[x]$ is he integral part of $x$ , and $x$ is its fractional part
How do I go about finding the value of $x(1-f)$ ?
Thanks!

Imagine expanding $A=(9+4\sqrt{5})^{48}+(9-4\sqrt{5})^{48}$ using the Binomial Theorem. The terms that involve odd powers of $\sqrt{5}$ cancel, so $A$ is an integer. 
Note that $(9-4\sqrt{5})^{48}$ is positive and close to $0$. It follows that the integer part of $(9+4\sqrt{5})^{48}$ is $A-1$, so the fractional part $f$ is $1-(9-4\sqrt{5})^{48}$. Now we have a nice expression for $1-f$. Multiply by $x$, using the fact that $(9+4\sqrt{5})(9-4\sqrt{5})=1$. 
Remark: The numbers $9+4\sqrt{5}$ and $9-4\sqrt{5}$ are conjugates. Whenever $9+4\sqrt{5}$ has a problem, $9-4\sqrt{5}$  can be reliably counted on for help.
