Fibonacci Polynomials!

• Let $ax^5 + bx^4 + 1 = Q(x)(x^2 - x - 1) + R(x)$ .
• We want $R(x)$ , the remainder $=0$
• Thus we set $(x^2 - x - 1) = 0$

$\begin{aligned} \implies x^2 &= x + 1 \\ x^{3} &= x^{2} + x^{} \quad {\color{#3D99F6}{\times x}} \\ x^{4} &= x^{3} + x^{2} \quad {\color{#3D99F6}{\times x}} \\ x^{5} &= x^{4} + x^{3} \quad {\color{#3D99F6}{\times x}} \end{aligned}$

$\begin{aligned} ax^5 + bx^4 + 1 &= a(x^4 + x^3) + bx^4 + 1 \\ &= (a+b)x^4 + ax^3 + 1 \\ &= (2a+b)x^3 + (a+b)x^2 + 1 \\ &= (3a+2b)x^2 + (2a+b)x + 1 \\ &= (5a+3b)x + (3a+2b) + 1 \\ \\ 0 &= (5a+3b)x + (3a+2b + 1) \end{aligned}$

\[\begin{cases} 5a + 3b = 0 \\ 3a + 2b + 1 = 0 \end{cases}

\implies a = 3, b = -5

\implies \boxed{a + b =-2}\]

For Fibonacci numbers $F$ :

$(x^2 - x - 1)((-1)^{n}F_{n}x^{n - 1} + (-1)^{n - 1}F_{n - 1}x^{n - 2} + ... -2x^2 + x - 1)$

$= ((-1)^{n}F_{n}x^{n + 1} + (-1)^{n - 1}F_{n - 1}x^{n} + (-1)^{n - 2}F_{n - 2}x^{n - 1} + ... -2x^4 + x^3 - x^2) + (-(-1)^{n}F_{n}x^{n} - (-1)^{n - 1}F_{n - 1}x^{n - 1} + ... -3x^4 + 2x^3 - x^2 + x) + (-(-1)^{n}F_{n}x^{n - 1} - (-1)^{n - 1}F_{n - 1}x^{n - 2} + ... + 5x^4 - 3x^3 + 2x^2 - x + 1)$

$= (-1)^{n}F_{n}x^{n + 1} + ((-1)^{n - 1}F_{n - 1} + (-1)^{n - 1}F_{n})x^{n} + ((-1)^{n - 2}F_{n - 2} + (-1)^{n - 2}F_{n - 1} - (-1)^{n - 2}F_{n})x^{n - 1} + ... + (-2 - 3 + 5)x^4 + (1 + 2 - 3)x^3 + (-1 - 1 + 2)x^2 + (1 - 1)x + 1$

$= (-1)^{n}F_{n}x^{n + 2} + (-1)^{n - 1}F_{n + 1}x^{n} + 1$

Therefore, $x^2 - x - 1$ divides $(-1)^{n}F_{n}x^{n + 1} + (-1)^{n - 1}F_{n + 1}x^{n} + 1$ .

In this question, $n = 4$ , so $x^2 - x - 1$ divides $F_{4}x^{5} - F_{5}x^{4} + 1 = 3x^{5} - 5x^{4} + 1$ , so $a = 3$ , $b = -5$ , and $a + b = \boxed{-2}$ .