Counting solutions to x^{p+1}=y^4 in a finite field - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T14:20:52Z http://mathoverflow.net/feeds/question/6186 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6186/counting-solutions-to-xp1y4-in-a-finite-field Counting solutions to x^{p+1}=y^4 in a finite field Nicholas Proudfoot 2009-11-19T22:25:16Z 2010-10-25T00:41:31Z <p>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.</p> <p>I feel as if this should be easy, and I'm missing an obvious approach. Can anyone help me out?</p> http://mathoverflow.net/questions/6186/counting-solutions-to-xp1y4-in-a-finite-field/6196#6196 Answer by Steven Sivek for Counting solutions to x^{p+1}=y^4 in a finite field Steven Sivek 2009-11-19T23:36:55Z 2009-11-19T23:36:55Z <p>Let g be a generator of the multiplicative group of the field; assuming x and y are nonzero, we can write x=g<sup>a</sup> and y=g<sup>b</sup> with 0 &lt;= a,b &lt; p<sup>2</sup>-1, and then x<sup>p+1</sup>=y<sup>4</sup> becomes g<sup>a(p+1)</sup>=g<sup>4b</sup>, or equivalently a(p+1) = 4b (mod p<sup>2</sup>-1).</p> <p>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 p<sup>2</sup>-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.</p> <p>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 p<sup>2</sup>-1, so there are 2(p-1)(p+1)+1 = 2p<sup>2</sup>-1 solutions.</p> <p>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 4p<sup>2</sup>-3 solutions.</p>