If $(x+1)^4+(x+3)^4=4$ then how to find the sum of non-real solutions of the given equation? If $(x+1)^4+(x+3)^4=4$ then how to find the sum of non-real solutions of the given equation?
I took $x+2=t$ and got $t^2=-6+\sqrt(40)/2$.How to proceed?

Put $t=x+2$ then you obtain $$(t-1)^4 +(t+1)^4 =4 .$$
hence $$((t-1)^2)^2 -2(t^2-1 )^2 +((t+1)^2)^2 =4 -2(t^2-1 )^2$$ hence $$16t^2 =4 -2(t^2-1 )^2$$ and let $s=t^2 -1 $ we obtain $$16 s +12 +2s^2 =0$$ therefore $$s^2 +8s +6 =0$$ thus $$x=-2\pm i\sqrt{3+\sqrt{10}}$$ or $$x=-2 \pm \sqrt{\sqrt{10}- 3}$$
