Question

Find integers 'a' and 'b' such that $(x^2-x-1)$ divides $a{x}^{17}+b{x}^{16}+1$.

Answer 1

Let $f\left(x\right)=a\cdot x^{17}+b\cdot x^{16}+1$

Given that $x^2-x-1$ divides $f\left(x\right)$, hence we can represent $f\left(x\right)$ as

$f\left(x\right)=(x^2-x-1)\cdot p\left(x\right)$ Eq.$\left(1\right)$

where $p\left(x\right)=c_1.x^{15}+c_2.x^{14}+...........c_{15}.x+c_{16}$ in which $c_1,c_2,...c_{16}$ are constant coefficients.

On comparing both sides of Eq.$\left(1\right)$, we see that $f\left(x\right)$ has its constant term as $+1$, which can only be achieved via multiplication of $x^2-x-1$ and $p\left(x\right)$, if and only if constant term $c_{16}=-1$

Now, $f\left(x\right)=(x^2-x-1)(c_1.x^{15}+c_2.x^{14}+...........c_{15}.x-1)$

As, $f\left(x\right)$ has no terms of $x$, we can say $c_{15}=1$

Similarly, on comparing the coefficients of $f\left(x\right)$ and the resultant from the multiplcation of $p\left(x\right)$ with $x^2-x-1$, we can easily evaluate coefficients $c_{14},c_{13}.....b,a$

On solving , we see that $c_{14}=-2,c_{13}=3,c_{12}=-5$ and so on

which gives a pattern with alternate + and - signs and with each next terms magnitude is equal to the sum of magnitude of last two terms. whiuch is the property for Fibbonaci series.

Therefore, for magnitude

$a=c_1=F_{17}$ and $b=c_1+c_2=-F_{16}$

where $F_{16}$ and $F_{17}$ are the sixteenth and seventeenth Fibbonaci numbers.

And sign we can adjust accordingly as the alternate sign is changing.

$\Rightarrow$ $a=F_{17}=1597$and $b=-F_{16}=-987$