Rational points on $ \frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0 $ , $k>3$, genus 0 - MathOverflow

Rational points on $ \frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0 $ , $k>3$, genus 0

2011-04-20T06:04:34Z

For integer $k>3$, is something known about the rational points on

$$ \frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0 $$

It is genus 0 curve for $3 < k < 100$.

Coprime integer solutions are unlikely because of the Fermat-Catalan conjecture.

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 $2^5=1^5+(2-1)^4 31$

Answer by Qiaochu Yuan for Rational points on $ \frac{ x^k-y^k }{ x-y } - (x-y)^{k-2} = 0 $ , $k>3$, genus 0

2011-04-20T06:25:43Z

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

$$(R : S) \mapsto \left( R(R - S)^{k-2} : S(R - S)^{k-2} : \frac{R^k - S^k}{R - S} \right).$$

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 $k-2$.)