A polynomial

The polynomial $q(x) = (x^2-1)p(x) - x$ has degree $2017$ and is such that $q(k) = 0$ for $2 \le k \le 2017$ . We know $2016$ distinct roots of $q(x)$ , and hence $q(x) \; = \; (a + bx)(x-2)(x-3)(x-4)\cdots(x-2017)$ for some constants $a,b$ . Since $\begin{aligned} -1 \; = \; q(1) & = (a+b) \times 2016! \\ 2 \; = \; 2q(-1) & = (a - b) \times 2018! \end{aligned}$ we deduce that $2016! a \; = \; -\frac{1017576}{2035153} \hspace{2cm} 2016! b \; = \; -\frac{1017577}{2035153}$ Thus $2017 \times 2019 p(2018) - 2018 \; = \; q(2018) \; = \; 2016!(a + 2018b)$ and so $p(2018) \; = \; \frac{504}{2035153}$ making the answer $\boxed{2035153}$ .

Relevant wiki: Polynomial Interpolation - Remainder Factor Theorem - Intermediate

Let $q(x)$ be a polynomial such that $q(x) = p(x) (x^2-1) -x$ . The polynomial $q(x)$ has degree $deg(p(x))+2 = 2014+2=2016$ (in fact $p(x)$ is multiplied by $x^2$ , so the degree is increased of 2).

Therefore we know exactly what $q(x)$ is. It is a polynomial of $2016$ th degree, so it has got at most 2016 roots. Since we have defined $q(x)$ as $q(x) = p(x) (x^2-1) -x$ , for every $k$ with $2 \le k \le 2017$ , $q(k) = 0$ . i.e. $q(230) = p(230)(230^2-1)-230 \rightarrow q(230) = \frac{230}{230^2-1} \cdot (230^2-1)-230 = 230-230=0$ .

We know exactly all the possible roots of $q(x)$ as we have found 2016 of them and they can be at most 2016. Thus,

$\large q(x)= (x-2)(x-3) \dots (x-2016)(x-2017)$

Then,

$\large q(2018)= (2018-2)(2018-3) \dots (2018-2016)(2018-2017) = 2016!$

Since the relation $q(x) = p(x) (x^2-1) -x$

$\large q(2018)= p(2018)(2018^2-1)-2018 \rightarrow p(2018)=\frac{2016!+2018}{2017 \cdot 2019}$

Now we need to find $b$ from the fraction $\frac{2016!+2018}{2017 \cdot 2019}$ . Rewrite it as:

$\large \frac{2016!+1+2017}{2017 \cdot 2019}$

From Wilson's theorem , $(p-1)! \equiv -1 \pmod p$ . Whereas 2017 is a prime number,

$\large 2016! \equiv -1 \pmod {2017} \rightarrow 2016!+1 \equiv 0 \pmod {2017}$ .

Therefore, $2017 | 2016!+1 \rightarrow 2017|2016!+1+2017$ . Now we have to verify if 2019 is the minimum denominator. The factorization of $2019$ is $3 \cdot 673$ , so we have to check if $3| \frac{2016!+2018}{2017}$ or $673| \frac{2016!+2018}{2017}$

• Since $(3,2017) = 1$ , $3| \frac{2016!+2018}{2017} \iff 3|2016!+2018$ . That's impossible because 2016! is divisible by 3, but 2018 is not divisible.

• Since $(673,2017) = 1$ , $673| \frac{2016!+2018}{2017} \iff 673|2016!+2018$ . That's also impossible as 2016! contains a 673 factor (so is divisible by 673), but 2018 is not divisible by 673.

So $p(2018) = \frac{a}{2019}$ with $a= \frac{2016!+2018}{2017}$ and $b = \boxed{2019}$