## Counting solutions to x^{p+1}=y^4 in a finite field

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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?

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"Reason to believe the answer is a single polynomial in p": What's the conjectured polynomial, or is there not an explicit one? That might help, and certainly couldn't hurt. (Well, it could hurt. But it's more likely to help.) – Harrison Brown Nov 19 2009 at 23:06
This is a small piece of a larger calculation in which I'm trying to understand the action of a particular finite group on the intersection cohomology of a particular complex variety.  The intersection cohomology is pure and concentrated in even degrees, so this computation is amenable to finite field techniques.  The fact that the answer to my question (as Sivek points out) DOES depend on the residue of p mod 4 means either that this dependence will be magically cancelled out by another part of the calculation, or that I've made a mistake earlier on. – Nicholas Proudfoot Nov 20 2009 at 0:23
The [finite-fields] tag is appropriate here! – Sonia Balagopalan Nov 21 2009 at 13:16

## 1 Answer

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.

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