Chain Reaction @ 02

We are given that $N(x^{2^{2008}-1}-1)=(x+1)(x^{2}+1)(x^{4}+1). . . . .(x^{2^{2007}}+1)-1$ . So, by multiplying by $x-1$ on both sides we get, $N(x-1)(x^{2^{2008}-1}-1)=(x^{2^{2008}}-1)-(x-1)$ $N(x-1)(x^{2^{2008}-1}-1)=(x^{2^{2008}}-x)$ $N(x-1)\frac{(x^{2^{2008}}-x)}{(x)}=(x^{2^{2008}}-x)$ $N=\large{\boxed{\frac{x}{x-1}}}$ . Now we are asked to find the number of solutions of the equation, $\left| \frac { x }{ x-1 } \right| =1$ CASE @ 1 $\frac { x }{ x-1 } =1$ $\Rightarrow$ No Solution.

CASE @ 2 $\frac{x}{x-1}=-1$ $\Rightarrow$ $\boxed{x=\frac{1}{2}}$ .

\therefore no. of solutions of |N|=1 is $\boxed{\boxed{1}}$