Anyone has a good idea? I'm very confused about this problem. I think the answer given in the Compendium is not right.

Please forgive me to replace some $\equiv$ with $=$. Notice that $f(z)=(z^4+2z^2-1)^2+32z^2$,and $p|f(z)$.We can conclude $p$ has quadratic residue $-1$,let $m^2=-1(modp),n^4=2(modp),q^4=-2(modp)$,then $((x-m)^4+q^4)((x+m)^4+q^4)=(x^2+1)^4+q^4((x-m)^4+(x+m)^4)+q^8=f(x)(modp)$. Take $z_1=m+n,z_2=m-n,z_3=m+mn,z_4=m-mn,z_5=-m+n,z_6=-m-n,z_7=-m+mn,z_8=-m-mn$,which could be easily checked satisfy the condition.Done

Do anyone know of a correct solution to this? I am not able to find any correct solution for this problem on the internet.