Rational points on $\frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0$ , $k>3$, genus 0 - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:13:48Z http://mathoverflow.net/feeds/question/62375 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62375/rational-points-on-frac-xk-yk-x-y-x-yk-2-0-k3-genus-0 Rational points on $\frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0$ , $k>3$, genus 0 jerr18 2011-04-20T06:04:34Z 2011-04-20T06:25:43Z <p>For integer $k>3$, is something known about the rational points on</p> <p>$$\frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0$$</p> <p>It is genus 0 curve for $3 &lt; k &lt; 100$.</p> <p>Coprime integer solutions are unlikely because of the Fermat-Catalan conjecture.</p> <p>Rational solutions with coprime numerators and denominator of suitable size appear unlikely because of possible abc triples related to $x^k = y^k+(x-y)^{k-1}$. (example with large denominator for $k=5$ is the point $(\frac{2}{31}, \frac{1}{31})$ with abc relation $2^5=1^5+(2-1)^4 31$</p> http://mathoverflow.net/questions/62375/rational-points-on-frac-xk-yk-x-y-x-yk-2-0-k3-genus-0/62377#62377 Answer by Qiaochu Yuan for Rational points on $\frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0$ , $k>3$, genus 0 Qiaochu Yuan 2011-04-20T06:25:43Z 2011-04-20T06:25:43Z <p>The projective closure is given by $\frac{X^k - Y^k}{X - Y} - Z(X - Y)^{k-2} = 0$. There is an obvious isomorphism from $\mathbb{P}^1$ given by</p> <p>$$(R : S) \mapsto \left( R(R - S)^{k-2} : S(R - S)^{k-2} : \frac{R^k - S^k}{R - S} \right).$$</p> <p>From here it is straightforward to parameterize the rational points on the affine curve. (Abstractly, this works because the tangent line at the origin intersects with multiplicity $k-2$.)</p>