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Let $f$ be an odd prime power and consider the Galois field of order $f$, $GF\left(f\right)$. Consider the equation $z^{2}=y^{k}+a_{1}y^{k-1}+a_{2}y^{k-2}+...+a_{k-1}y+a_{k}$, where $\lbrace z,y,a_{1},...,a_{k} \rbrace$ are elements of $GF\left(f\right)$. what is the number of solutions $(z,y)$, $z\neq 0$, that satisfy in this equation?

FBJ
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    You're asking for the number of points on a curve over a finite field. The exact number depends on the exact curve, but there is are useful estimates around. Look into the work of Weil on the question. Or get a copy of the textbook by Ireland and Rosen. – Gerry Myerson Nov 19 '16 at 11:55
  • To build on Gerry's comment: If $N$ is the number of solutions, then, in the generic case, Weil's bound gives an estimate of the type $$|N-(f+1)|\le (k-1)\sqrt f.$$ Some adjustments may be needed re the points at infinity (depending on the details of the polynomial on the r.h.s), but you should expect the approximate interval $f\pm (k-1)\sqrt f$. There are some degenerate cases, where this doesn't hold. Such as $z^2=p(y)^2$, $p$ a polynomial of degree $k/2$, when we get approximately $2f$ solutions. In such a case the underlying curve is not irreducible, and the theory doesn't apply. – Jyrki Lahtonen Nov 20 '16 at 06:52

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