Find the value of $x$ which is correct I have one exercise which is $$(x+2013)(x+2014)(x+2015)(x+2016)+1=0$$
I tag $A=x+2013$ or other for many ways but still can not find the first $x$ value. please help.

If you let $y=x+2014$, then the equation becomes
$$(y-1)y(y+1)(y+2)+1=0 \Leftrightarrow (y^2+y-1)^2=0$$
So
$$y = \frac{-1\pm\sqrt{5}}{2} \Rightarrow x = \ldots$$

\begin{align}
(y-1)y(y+1)(y+2)+1 & = y^4+2 y^3-y^2-2 y+1 \\
{} & = \left( y^4+y^3-y^2 \right) + y^3-2y+1 \\
{} & = y^2 \left( y^2+y-1 \right) + \left(y^3+y^2-y\right) -y^2-y+1 \\
{} & = y^2 \left( y^2+y-1 \right) + y\left(y^2+y-1\right) -\left( y^2+y-1\right) \\
{} & = \left( y^2+y-1 \right)^2
\end{align}

\begin{align}
(y-1)y(y+1)(y+2)+1 & = y^4+2 y^3-y^2-2 y+1 \\
{} & = \left( y^4+2 y^3+y^2 \right) - 2y^2-2 y+1 \\
{} & = \left( y^2 + y \right)^2 - 2(y^2+y)+1 \\
{} & = \left( y^2+y-1 \right)^2
\end{align}
