# Rational points on the “quintic circle” $x^5 + y^5 = 7$

I suspect that the curve $x^5 + y^5=7$ has no $\mathbb Q$ points, and a brief computer search verifies this hypothesis for denominators up to $10^4$. What techniques can be used to show that there are no solutions?

• It helps to instead consider the corresponding projective curve $X^5 + Y^5 = 7Z^5$ in $\mathbb{P}^2$ of genus $6$. This has a rational point, namely $(X:Y:Z) = (1:-1:0)$. Your conjecture is that this is the only rational point. The similarity here to the $n=5$ case of Fermat's last theorem should be very clear. Of course the proof of FLT for all $n$ was very hard, however there exist elementary proofs for the $n=5$ case (en.wikipedia.org/wiki/…). Perhaps these could be adapted to your case. – Daniel Loughran Nov 22 '15 at 8:58
• The modular techniques and Frey curve machinery used to prove FLT could also possibly be made to apply in your case, but I am not an expert in these methods. – Daniel Loughran Nov 22 '15 at 9:03

There is an action of $\mu_5$, the group of fifth roots of unity, on your curve, given by $\zeta \cdot (x,y) = (\zeta x, \zeta^{-1} y)$. The quotient by this group action is the hyperelliptic curve $$C \colon Y^2 = X^5 + \frac{49}{4},$$ the map being given by $(X, Y) = (-xy, x^5 - \frac{7}{2})$. So it is enough to find all the rational points on $C$. $C$ is isomorphic to $C' \colon Y^2 = 4 X^5 + 49$. By a 2-descent, one can show that the Jacobian variety of $C'$, $J'$, has Mordell-Weil rank (at most) 1, and since one finds a point of infinite order ($(x^2 - \frac{10}{9} x - \frac{10}{9}, \frac{200}{27} x - \frac{61}{27})$ in Mumford representation), Chabauty's method is applicable and shows that $\infty$, $(0, \pm \frac{7}{2})$ are the only rational points on $C$. This implies that there are no rational points on your affine curve.

Here is Magma code to check this:

P<x> := PolynomialRing(Rationals());

C := HyperellipticCurve(4*x^5 + 49);

J := Jacobian(C);

RankBound(J); // --> 1

ptsJ := Points(J : Bound := 500); // the first 5 are torsion

Chabauty(ptsJ[6]); // --> { (0 : -7 : 1), (1 : 0 : 0), (0 : 7 : 1) }

(If you do not have direct access to Magma, you can try this out with the online Magma calculator at http://magma.maths.usyd.edu.au/calc/ .)

• Could you explain 'By a 2-descent, one can show that the Jacobian variety of C′'? – user76479 Nov 22 '15 at 12:31
• @Arul : This is explained in detail in my paper "Implementing 2-descent for Jacobians of hyperelliptic curves" (Acta Arith. 98, 245-277, 2001). If you are familiar with 2-descent on elliptic curves, then it can be seen as a generalization of that: one computes the 2-Selmer group of the Jacobian as a subgroup of ${\mathbb Q}(\theta)^\times$ mod squares, where $\theta = \sqrt[5]{49/4}$. There is an injection of $J({\mathbb Q})/2 J({\mathbb Q})$ into the 2-Selmer group, so we obtain an upper bound for the rank. – Michael Stoll Nov 22 '15 at 12:42
• It is worth pointing out that passage from $x^5+y^5=7$ to the hyperelliptic curve is achieved by rewriting as $x^{10}+(xy)^5=7x^5$ changing variables to $u=x^5$ and $v=xy$ and completing the square. – pre-kidney Jun 30 at 3:52

In his 1825 paper, Lejeune Dirichlet proved that the equation $x^5 + y^5 = cz^5$ has no nontrivial solution with x and y coprime integers and z integer for a rather large class of integers c. His paper can be read on the site Gallica : http://gallica.bnf.fr/ark:/12148/bpt6k5842885h/f2.item.zoom but the visualization is very bad.

• It's actually not so bad already, but here is an improved OCRed file: dropbox.com/s/jkbl5gtzdyuzktt/dirichlet.pdf?dl=0 – Igor Rivin Nov 22 '15 at 12:32
• Looking through Dirichlet's paper, it seems that he always assumes $c$ to be divisible by 2 or 5 (unless $c = 1$). – Michael Stoll Nov 22 '15 at 12:37
• You are right. Dirichlet doesn't solve the problem for $c = 5$. My answer was rather a comment. – Panurge Mar 20 '16 at 10:57
• @Panurge meant to write "Dirichlet doesn't solve the problem for $c=7$". – pre-kidney Jun 30 at 3:46