Is it possible? 2

Lemma: If $P(x)$ is a polynomial with integer coefficients, then $x-y | P(x)-P(y)$ for every $x$ , $y$ integer.

Proof: Letting $P(x) = \displaystyle \sum_{i=0}^n a_ix^i$ with $a_i$ being an integer for every $i$ we obtain

$P(x)-P(y) = \displaystyle \sum_{i=0}^n a_ix^i - \displaystyle \sum_{i=0}^n a_iy^i = \displaystyle \sum_{i=0}^n a_i(x^i-y^i)$

Clearly $x-y | x^i-y^i$ for $i \geq 0$ (since $x^i-y^i = (x-y)(x^{i-1}+x^{i-2}y+...+xy^{i-2}+y^{i-1})$ for $i>0$ and $x^i-y^i = 1-1 = 0$ for $i=0$ ), which means that

$x-y | \displaystyle \sum_{i=0}^n a_i(x^i-y^i) = P(x)-P(y)$ .

Applying the lemma for the polynomial described in the problem with $x=2018$ and $y=2015$ we would have

$3 = x-y | P(x)-P(y) = 1-2016 = -2015$

but $3 \nmid (-2015)$ so there does not exist such a polynomial.