[math-fun] square roots in GF(2^k) ?

I did a quick Google search on discrete square roots, but most refs were for GF(p). Is the complexity of taking a square root in GF(2^k) any less than taking an arbitrary discrete log? How about the simple question of whether an element *has* a square root? The reason I ask this question is due to the fact that: (a+b)^2 (mod pp) = a^2 + 2ab + b^2 = a^2+b^2 (mod pp), where pp is the irreducible polynomial for the field GF(2^k), and a,b are representative polynomials whose degree is < deg(pp)=2^k. So, (a*x^j+b)^2 = (a*x^j)^2 + b^2 = a^2*x^2j + b^2 So Karatsuba is even simpler/faster in these fields, hence, it might seem that finding such a square root might be easier than in some other fields.