jmc

Since $x^{2017}+Ax+B$ is divisible by $(x+1{)}^2,$ it must have $x=-1$ as a root, so $$(-1{)}^{2017}+A(-1)+B=0,$$or $A=B-1.$ Then $x^{2017}+Ax+B=x^{2017}+(B-1)x+B.$ Dividing this polynomial by $x+1$, we have $$\begin{array}{rl}\frac{x^{2017}+(B-1)x+B}{x+1}& =\frac{x^{2017}+1}{x+1}+(B-1)\\ & =(x^{2016}-x^{2015}+x^{2014}+\cdots +x^2-x+1)+(B-1),\end{array}$$which must be divisible by $x+1.$ Therefore, setting $x=-1,$ we get $$\left((-1{)}^{2016}-(-1{)}^{2015}+(-1{)}^{2014}+\cdots +(-1{)}^2+1\right)+(B-1)=0,$$or $B+2016=0.$ Thus, $B=\boxed{-2016}.$