Functional Remainder

Let $\mathcal{P(x)}=x^{999}$ and let the quotient when $\mathcal{P(x)}$ is divided by $x^2-4x+3$ be $\mathcal{Q(x)}$ .

Also note that $x^2-4x+3=(x-1)(x-3)$

Therefore, we can write this division as $\large \mathcal{P(x)}=x^{999}=\mathcal{Q(x)}(x-1)(x-3)+ax+b$

Where $ax+b$ is the remainder for some integers $a$ and $b$ .

Now, $\mathcal{P(1)}=a+b=1$ . ---(1)

Similarly, $\mathcal{P(3)}=3a+b=3^{999}$ . ---(2)

On solving these two equations , we get $a=\dfrac{3^{999}-1}{2}$ and $b=\dfrac{3-3^{999}}{2}$ .

$\implies$ Remainder is $\boxed{\dfrac{3^{999}-1}{2}x+\dfrac{3}{2}(1-3^{998})}$