Hints: Show that $\frac{1}{x}$ is an integer. Then show that $|x|=1$.

$1=2008-2007=x^{n}+tx-x^{n}-sx=(t-s)x \implies x=\frac{1}{t-s}$ $\implies 2008=(\frac{1}{t-s})^n+\frac{t}{t-s} \implies 2008(t-s)^n=t(t-s)^{n-1}+1$ $\implies t-s|1 \implies t-s=1,-1 \implies x=\frac{1}{t-s}=1,-1$ $ i) x=1\implies 2008=x^n+tx\implies\boxed{t=2007}$ , $2007=x^n+sx \implies\boxed{s=2006}$ $ii)x=-1,n=2k\implies 2008=x^n+tx\implies\boxed{t=-2007} $ , $2007=x^n+sx\implies\boxed{s=-2006}$ $iii)x=-1,n=2k+1\implies 2008=x^n+tx\implies\boxed{t=-2009}$ , $2007=x^n+sx\implies\boxed{s=-2008}$

clearly tx-sx=1 i.e. (t-s)x=1 ,so x =1 or -1 . if x=1 , then consider he case n=even then s=2006 , and t=2007. similarly , other cases can be dealt

mathbuzz wrote: clearly tx-sx=1 i.e. (t-s)x=1 ,so x =1 or -1 if $x\in Z$ its true , but if $x\in R$ it can be wrong. for example: $x=\frac{1}{2},(t-s)=2$ $\implies$ we need to other cases for prove that :{$x = 1$ or $-1$}