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This might be a counterexample to a conjecture of Granville about automorphisms of twists of hyperelliptic curves.

In this paper,

the quadratic twist of $f(x)=y^2$ is denoted by $C_d : d y^2=f(x)$ on p.1. where $f(x)$ is squarefree.

From p. 3 We denote by $Aut(C)$ the (finite) group of such automorphisms of $C$, and define $c_d(\mathbb{Z})$ and $c_d (\mathbb{Q})$ to be the number of automorphism classes of non-trivial integral and rational points on $C_d$ , respectively.

Conjecture 2 (ii) There are only finitely many squarefree integers $d$ for which $c_d (\mathbb{Q}) > 2$ when $\deg(f)>d_0$.

Proposed counterexample which stands numerical experiments.

Let $a(x)=(-x^2+5x+1)^2,b(x)=(-x^2-5x+1)^2$ and $f(x)=a(x)^n+b(x)^n$ for even natural $n$.

$f(x)$ is even and symmetric, so we have the automorphisms $(x,y) \mapsto (-x,y), (x,y) \mapsto (1/x,1/x^{2n})$ for all $d$.

Let's solve $a(x_1)=A,b(x_1)=B,a(x_2)=Bu,b(x_2)=Au$ over the rationals.

Let $D=numerator(A^n+B^n)$. Write $D=d D'^2$ with $d$ squarefree. We have $f(x_1),f(x_2)$ on $C_d$ since $f(x_1)=D=d D'^2$ and $f(x_2)=Du^n=d u'^n$ ($n$ is even).

Solving symbolically gives three solutions. The first two are the known automorphisms and the third is

$x_2=1/2/(-1+x_1^2)(-25x_1+(y_1 ))$ for $x_1,y_1$ on $C : 617x_1^2+4x_1^4+4-y_1^2=0$

$C$ has infinitely many rational points, which gives the third solution on $C_d$ whenever $x_1,y_1$ are on $C$.

So third candidate automorphism is $(x_1,y) \mapsto (1/2/(-1+x_1^2)(-25x_1+(y_1 )),\ldots)$.

To show the candidate is automorphism, one must show $x_2$ is $x_1$ coordinate on $C$, which is experimentally true (and this is independent of $n$).

We tried to showed the above, but didn't succeed.

Q1 Can this argument be made rigorous?

As an aside, this gives infinitely many rational and integer non-trivial solutions to $(a(x_1)^n+b(x_1)^n)(a(x_2)^n+b(x_2)^n)=y^2$.

Sage verification, code, might be run in the cloud:

def granvsym2():
    K.<x>=QQ[]
    #on 617*x1^2+4*x1^4+4-y1^2=0
    n=8 #change to arbitrary large even
    E=EllipticCurve(QQ,[0, 0, 0, -380881/3, 469059442/27])
    P=E.gens()[0]
    h=(-x^2-5*x+1)^(2*n)
    f=(-x^2+5*x+1)^(2*n) + h
    g=(617*x^2+4*x^4+4)
    for k in [ 2 .. 10]:
        u,v=(k*P).xy()
        u= -u
        x1,y1=-36/(380113+3702*u+9*u^2)*v, -1/2*(-7611476-29616*u+36*u^2)/(380113+3702*u+9*u^2)
        x2=1/2/(-1+x1^2)*(-25*x1+(y1 ))
        print k,QQ(f(x1)/f(x2)).is_square(),g(x1)-y1^2==0,'---','x2squ=',QQ(g(x2)).is_square()
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  • $\begingroup$ According to Granville, an "automorphism class" is an orbit of the automorphism group acting on rational points, not an automorphism itself. It seems like you are constructing families of hyperelliptic curves with lots of automorphisms, which does not have any bearing on Granville's conjecture. $\endgroup$ Aug 4, 2015 at 16:11

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