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.

HINT:
Set $y=\dfrac{x+2013+(x+2014)+(x+2015)+(x+2016)}4$
$$\iff x+2013=y-\dfrac32$$
$$0=(4y^2-9)(4y^2-1)+16=(4y^2-5)^2$$
