Quadratic EquationHard

Question

If the value of quadratic trinomial $ax^{2} - bx + c$ is an integer for $x = 0,x = 1$ and $x = 2$, then the valueof the given trinomial is an integer for

Options

A.$x = 2017$
B.$x = 2018$
C.$x = - 4$
D.$x = - 2017$

Solution

$f(x) = ax^{2} - bx + c$

$$\begin{matrix} & f(0) = c \in I \\ & f(1) = a - b + c = k_{1} \\ & f(2) = 4a - 2\text{ }b + c = k_{2} \end{matrix}\ k_{1},k_{2} \in I$$

$$\begin{matrix} a - b & \ \in I \\ 2a + 2(a - b) + c & \ = k_{2} \\ 2a & \ \in I\ \Rightarrow \ 2\text{ }b \in I \\ f(2k) & \ = 4ak - 2bk + c \in I \\ f(2k + 1) & \ = a(2k + 1)^{2} - b(2k + 1) + c = 4{ak}^{2} + 4ak + a - 2bk - b + c \\ & \ = 4{ak}^{2} + 4ak - 2bk + (a - b + c) \in I \\ \Rightarrow \ f(x) & \ \in I\forall x \in I \end{matrix}$$

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