Note that $p\triangleq 2017$ is a prime. We rewrite the expression as \[ p(a+b)=(b-a)(a^2+ab+b^2). \]Case 1. $a+b$ is odd. Note that $(a+b,b-a)=1$. For this reason, $b-a\mid p$, yielding $b-a\in\{1,p\}$. If $b-a=p$ then $a+b=a^2+ab+b^2>a+b$, a contradiction. Likewise if $b-a=1$ then $p(a+b) = (a^2+ab+b^2)\implies p(2a+1) = (3a^2+3a+1)$. In particular, $2a+1\mid 3a^2+3a+1 \implies 2a+1\mid 6a^2+6a+2$. Moreover, $2a+1\mid 6a^2+3a$, yielding $2a+1\mid 3a+2\implies 2a+1\mid 6a+4$. Likewise $2a+1\mid 6a+3$, both of which collectively yield a contradiction. Case 2. $a+b$ is even. In this case, $(a+b,b-a)=2$. Hence, $(b-a)/2\mid p$. Thus $b-a = 2$ or $2p$. Both cases are handled analogously.