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x^2+1 has a root in Zp, is equivalent to having an element of order 4 in Zp^x

First, what is Zp

**^x**,

Second are we considering the order under the multiplicative operation. So in this case, if I consider Z5, then no element has order 4 yet x= 2, and x =3 are both solution for x^2+1 !!

thanks