Perplexing Years!

Making $x=2008$ . So $P=x^{2007}-x$ and $Q=x^{2}+x+1$ . Thus, we have to solve $x^{2007}-x \pmod{x^{2}+x+1}$ . How $x^{2}+x+1\equiv 0\pmod{x^{2}+x+1}$ $\Rightarrow$ $x^{2}\equiv -(x+1)\pmod{x^{2}+x+1}$ $\Rightarrow$ $(x^{2})^{2}\equiv (-(x+1))^{2}\pmod{x^{2}+x+1}$ $\Rightarrow$ $x^{4}\equiv x^{2}+{2x}+1=x^{2}+x+1+x\equiv x\pmod{x^{2}+x+1}$ . That is, $x^{4}\equiv x\pmod{x^{2}+x+1}$ and $gcd(x,x^2+x+1)=1$ $\Rightarrow$ $x^{3}\equiv 1\pmod{x^{2}+x+1}$ $\Rightarrow$ $x^{2007}\equiv 1\pmod{x^{2}+x+1}$ $\Rightarrow$ $x^{2007}-x\equiv 1-x\pmod{x^{2}+x+1}$ . Therefore, $2008^{2007}-2008\equiv 1-2008=-2007\pmod{2008^{2}+2009}$ . How $2008^{2}+2009=4034073$ , so $2008^{2007}-2008\equiv -2007\equiv 4034073-2007=\boxed{4032066}\pmod{2008^{2}+2009}$ .

$2008^3=(2008^2+2009)+1$ $2008^{2007}=K \times (2008^2+2009)+1$ $Res = 2008^2+2009 - 2007 = \boxed{4032066}$

Let $x=2008$ . Note that since $2007$ is divisible by $3$ , $x^{2007}-1$ is divisible by $x^3-1$ , so for some polynomial $P(x)$ with integer coefficients we have $\begin{aligned}\dfrac{x^{2007}-x}{x^2+x+1}&=\dfrac{x^{2007}-1}{x^2+x+1}+\dfrac{1-x}{x^2+x+1}\\&=\dfrac{(x^3-1)P(x)}{x^2+x+1}+\dfrac{1-x}{x^2+x+1}\\&=(x-1)P(x)+\dfrac{1-x}{x^2+x+1}.\end{aligned}$ The first term is necessarily an integer, so the remainder is just $1-x=-2007\equiv 2008^2+2\equiv\boxed{4032066}\pmod{2008^2+2009}.$

Let x = 2008

x^2007 - x

= x^2007 + x^2006 + x^2005 - x^2006 - x^2005 - x

= x^2005 (x^2 + x + 1) - (x^2006 + x^2005 + x^2004) + x^2004 - x

= x^2005 (x^2 + x + 1) - x^2004 (x^2 + x + 1) + x^2004 - x

--> x^2004 - x

--> x^(3n + 0) - x {2007/ 3 = 669 and 2004/ 3 = 668}

--> 1 - x or x^3 - x

Simplified positive numbers = (x3 - x)/ (x2 + x + 1) = x (x2 - 1)/ (x2 + x + 1)

= 2008 * 4032063/4034073

= 8096382504/ 4034073

Remainder = MOD(8096382504, 4034073) = 4032066