I need to compute the number of solutions to the equation $x^{p+1} = y^4$ in the field with $p^2$ elements (for p sufficiently large). The form of the equation suggests to me that the solution would depend on the congruence class of p mod 4, but I have reason to believe that the answer is a single polynomial in p.


I feel as if this should be easy, and I'm missing an obvious approach. Can anyone help me out?



Counting solutions to x^{p+1}=y^4 in a finite field
Let g be a generator of the multiplicative group of the field; assuming x and y are nonzero, we can write x=ga and y=gb with 0 <= a,b < p2-1, and then xp+1=y4 becomes ga(p+1)=g4b, or equivalently a(p+1) = 4b (mod p2-1).


From this we see that p+1 | 4b is necessary, and if 4b=k(p+1) then (a,b) gives a solution iff a=k (mod p-1). Since a can range from 0 to p2-2, then, there are either 0 solutions or p+1 solutions for any fixed b. The total number of nonzero solutions is therefore (p+1)\* #{b | p+1 divides 4b}, and then (x,y)=(0,0) is the remaining solution.


Now if p is 1 (mod 4) we have p+1 | 4b iff b is a multiple of (p+1)/2, and there are 2(p-1) such b up to p2-1, so there are 2(p-1)(p+1)+1 = 2p2-1 solutions.


On the other hand, if p is 3 (mod 4) then p+1 | 4b iff b is a multiple of (p+1)/4, so we have 4(p-1) such b and there are 4p2-3 solutions.


