More fun in 2015, Part 5

Our friend Brian has written a fine solution. For the sake of variety, let me show how I thought of it as I wrote the problem.

Let $\phi=\frac{1+\sqrt{5}}{2}$ , the Golden Section, one of the roots of $x^2-x-1$ ; thus we have $\phi^2=\phi+1$ . We can show by induction that $\phi^n=F_n\phi+F_{n-1}$ , where $F_n$ is the nth Fibonacci number. The equation is trivial for $n=1$ , and $\phi^{n+1}$ $=F_n\phi^2+F_{n-1}\phi$ $=F_n(\phi+1)+F_{n-1}\phi=(F_n+F_{n-1})\phi+F_n$ $=F_{n+1}\phi+F_n$ .

Writing $x^n=q(x)(x^2-x-1)+ax+b$ and substituting $x=\phi$ , we find $\phi^n=F_n\phi+F_{n-1}=a\phi+b$ , so $a=F_n$ , $b=F_{n-1}$ , and $a+b=F_{n+1}.$

Now $b^2+ab-a^2=b(a+b)-a^2=F_{n-1}F_{n+1}-F_n^2=(-1)^n$ by Cassini's Identity. For the odd number $n=2015$ the answer is $\boxed{-1}$ .

This is not the ideal solution but I thought I'd get something posted anyway.

I started by dividing $x^{2015}$ by $x^{2} - x - 1$ using the "old-fashioned" long division method and found that the leading and trailing coefficients in the $n$ th division step were $F_{n+1}$ and $F_{n},$ respectively, (where $F_{k}$ is the $k$ th Fibonacci number with $F_{1} = F_{2} = 1$ ). Since there are $2014$ division steps before getting a remainder of the form $ax + b,$ we end up with $a = F_{2015}$ and $b = F_{2014}.$

So now we have $b^{2} + ab - a^{2} = F_{2014}^{2} + F_{2015}F_{2014} - F_{2015}^{2} =$

$F_{2014}(F_{2014} + F_{2015}) - F_{2015}^{2} = F_{2014}F_{2016} - F_{2015}^{2} = (-1)^{2015} = \boxed{-1},$

where we have invoked Cassinin's identity .