Remainder

Suppose $(x^2+1)(x^2+x+1) = 0$ . Then either $x^2+1 = 0$ or $x^2 + x + 1 = 0$ . Now we note:

• If $x^2+1=0$ , then $x^4 = 1$ .
• If $x^2+x+1 = 0$ , then $x^3 - 1 = (x-1)(x^2+x+1) = 0$ , so that $x^3 = 1$ .

In either case, we additionally see that $x^{12} = 1$ . In other words, every root of $(x^2+1)(x^2+x+1)=0$ satisfies $x^{12}-1=0$ . Since we note the former only has simple zeros, we can conclude $(x^2+1)(x^2+x+1)$ divides $x^{12}-1$ . Therefore, we have shown that $x^{12} \equiv 1 \pmod{(x^2+1)(x^2+x+1)}.$

Using this, we may write $x^{1959}-1 = (x^{12})^{163} \cdot x^3 - 1 \equiv x^3 - 1 \pmod{(x^2+1)(x^2+x+1)}$

Luckily, this is already reduced! Therefore, $a=1, b=0, c=0, d=-1$ and $a+b+c+d = \boxed{0}$ .

Let $(x^2 + 1)(x^2 + x + 1) = k$

$x^{1959} - 1\equiv{-x - 1}\equiv{x(x - 1)}\pmod{x^2 + 1}$

Multiplying throughout by $x^2 + x + 1$ yields ,

$(x^{1959} - 1)(x^2 + x + 1)\equiv{x(x - 1)(x^2 + x + 1)}\equiv{x(x^3 - 1)}\pmod{k}$ $\quad \quad \quad \cdot \cdot \cdot (A)$

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Now , $x^2 + x + 1 \mid x^3 - 1 \mid (x^{3})^{653} - 1^{653} = x^{1959} - 1$

$\Rightarrow x^{1959} - 1\equiv{0}\pmod{x^2 + x + 1}$

Multiplying throughout by $x^2 + 1$ gives ,

$(x^{1959} - 1)(x^2 + 1)\equiv{0}\pmod{k}$ $\quad \quad \quad \cdot \cdot \cdot (B)$

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Combining $(A) , (B)$ gives ,

$(x^{1959} - 1)(x)\equiv{x(x^3 - 1)}\pmod{k}$

$\Rightarrow (x^{1959} - 1)\equiv{x^3 - 1}\pmod{k}$

Hence , $a = 1 , b = 0 , c = 0 , d = -1 \Rightarrow a + b + c + d = \boxed{0}$