I've seen it before - 8

Let $\alpha$ and $\alpha^2$ be the roots of $f(x)=0$ . So, we have $f(\alpha)=0$ $\alpha^2-\alpha=k\rightarrow\text{Eq.1}$ By Vieta's formulae, we get $\alpha^2+\alpha=1\rightarrow\text{Eq.2}$ Eq.1+Eq.2 $2\alpha^2=1+k$ $\alpha^2=\dfrac{1+k}{2}\rightarrow\text{Eq.3}$ Eq.1-Eq.2 $\alpha=\dfrac{1-k}{2}\rightarrow\text{Eq.4}$ From Eq.3 & Eq.4, we get $\dfrac{1+k}{2}=\left(\dfrac{1-k}{2}\right)^2$ Expanding the above expression, we get $k^2-4k-1=0$ $\boxed{k=2\pm\sqrt{5}}$ Now we need to find $2^{2014}\mod 25$

We know that $\phi(25)=20$ ,so by Euler-Fermat theorem, we have $2^{20}\equiv1\mod125$ Raise both sides to 100th power $2^{2000}\equiv1\mod125\rightarrow Cg.1$ and also $2^7\equiv128\equiv 3\mod 25$ $2^{14}\equiv 9\mod 25 \rightarrow Cg.2$ Multiplying both the congruences, we get $\boxed{2^{2014}\equiv9\mod25}$

The roots of $f(x) = x^2 - x -k$ are $\dfrac {1 \pm \sqrt{1+4k}} {2}$ .

Assuming that $\dfrac {1 + \sqrt{1+4k}} {2} = \left( \dfrac {1 - \sqrt{1+4k}} {2} \right)^2$

$\Rightarrow \dfrac {1 + \sqrt{1+4k}} {2} = \dfrac {1 - 2\sqrt{1+4k} + 1 + 4k} {2} = \dfrac {1 + 2k - \sqrt{1+4k}} {2}$

$\Rightarrow 1 + \sqrt{1+4k} = 1 + 2k - \sqrt{1+4k} \quad \Rightarrow \sqrt{1+4k} = k \quad \Rightarrow 1+4k = k^2$

$\Rightarrow k^2 - 4k -1 = 0 \quad \Rightarrow k = \dfrac {4 \pm \sqrt {16+4} } {2} = 2 \pm \sqrt{5} \quad \Rightarrow a = 2, \quad b =5$

We need to find $x$ in:

$2^{2014} \equiv x \pmod {25} \equiv 2^{4}(1024)^{201} \pmod {25} \equiv 2^{4}(-1)^{201} \pmod {25}$

$\equiv 2^{4}(-1) \pmod {25} \equiv -16 \pmod {25} \equiv \boxed{9} \pmod {25}$