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$p(x) = x^{\frac{N}{2}}$
we ask what is $p(\zeta^{5^j})$
for $j=0$ we have $p(\zeta^{5^j}) = p(\zeta) = \zeta^{\frac{N}{2}} = i$
the rest are $i^{5^j}$. and $i^1=i, i^3=-i$ 
