If I haven't made any mistakes, the following proves that there are no rational points to the above equation[Complete revamp of answer. It is based on the one before, but is better!]
Let's start by makingIn Tim's hyperelliptic model a bit nicer. Changeequation, make the change of variables $y$ to $y/(-7)^2$, and $x$ to $x/(-7)$. Multiply by the common denominator $7^4$, to get:
$$y^2=x(x+7)(x^3+56x^2+245x-343)$$
For anyevery prime, if $v_p(x)<0$ or $v_p(y)<0$ then $2v_p(y)=5v_p(x)$. So change coordinates to$p$ with $(\frac{x}{z^2}, \frac{y}{z^5})$$v_p(x)<0$, and multiply outthe valuation must in fact be even, thus appears to get:
$$y^2=x(x+7z^2)(x^3+56x^2z^2+245xz^4-343z^6)$$
So $x,y,z$ are integers with $(x,z)=(y,z)=1$.
Now we consideran even power in the gcd'sfactorisation of each of the factors on the right:
$$(x,x^3+56x^2z^2+245xz^4-343z^6) | 343$$
$$(x+7z^2,x^3+56x^2z^2+245xz^4-343z^6) | 343$$
The second follows from reducing the cubic mod $x+7z^2$(as non-zero rational numbers).
[EDIT] for now, assume all three factors are positive. We will treatneed only consider the other cases below.
Hence $x^3+56x^2z^2+245xz^4-343z^6$ is either a square, or $7$ times a square. The first case is impossible since sage says thatsquarefree parts of the equation
$$y^2=x^3+56x^2+245x-343$$
has no rational pointsfactors.
If it is Assume $7$ times a square$p$ divides the numerator of at least two of the factors, then either $7|x$ or $7|x+7z^2$to an odd power. In either caseBy a small gcd calculation $7|x$(3 in fact), which implies $49|y$. Setwe see that $y=49y'$,$p$ must be $x=7x'$:
$$y'^2=7x'(x'+z^2)(x'^3+8x'^2z^2+5x'z^4-z^6)$$$7$.
And the cubicThus, we must have that each factor is a square ($n^2$). Since $(x,x+7z^2)|7$, we havetimes a number in $(x',x'+z^2)=1$$\{ -7, -1, 1, 7 \}$. Hence either $7|x'$ orFor each triple $7|x'+z^2$. Assume$(a,b,c)$ of numbers in the latterset, so thatwith $x'$ is$abc$ a square, and $x'+z^2$there is a sumpossible element of squares and also divisiblethe 2-Selmer group defined by 7. This implies $7|x'$ andthe curve $7|z$, contradicting$C_{a,b,c}$ $(x',x'+z^2)=1$.
We are left with checking the possibility that(in $7|x'$. But this would mean that
$$n^2=x'^3+8x'^2z^2+5x'z^4-z^6 \equiv -z^6\pmod{7}$$$\mathbb{P}^4$):
which is impossible, since it implies$$x = au^2,$$ $7|z$, once again contradicting$$\\ x+7=bv^2,$$ $(x',x'+z^2)=1$.
Now we treat negative numbers.$$x^3+56x^2+245x-343=cw^2$$
Hence $x^3+56x^2z^2+245xz^4-343z^6$ is either $-$square, or $-7\times$square. Sage says that the first case isn't possible. Also, exactly one of the two linear factors must be negative, which is impossible if $x$ is positive. Make the same changeSome of variables as above to get:
$$y'^2=7x'(x'+z^2)(x'^3+8x'^2z^2+5x'z^4-z^6)$$
these might not have points locally and we knowwill not define an element of the cubic is now $-$square and $x'$ is negative. There are two cases (last branch, I promise): $x'=-7u^2$ or $x'=-u^2$, for some non2-zero $u\in \mathbb{Z}$Selmer. But we will not make any explicit local computations.
In the first caseFor each such triple, for somethe curve $y_0$ we have:
$$y_0^2 = 7x'(x'^3+8x'^2z^2+5x'z^4-z^6)$$
By making$C_{a,b,c}$ has a changemorphism into each of variables with $x_0=-7z^2/x'$, there's a $y_0$ withthe curves:
$$y_0^2=x_0^3+35x_0^2-392x_0+343$$$$C_{a,b,c}^1:\\ y^2=c(x^3+56x^2+245x-343)$$
$$C_{a,b,c}^2:\\ y^2=acx(x^3+56x^2+245x-343)$$
$$C_{a,b,c}^3:\\ y^2=bc(x+7)(x^3+56x^2+245x-343)$$
ButA sage sayscomputation shows that this equationfor each such triple $(a,b,c)$, at least one of these three curves has no rational points.
The second and final case: $x'=-u^2$, $x'+z^2=7v^2$(other than ones at infinity, soor points that for some $y_0$ and $x_0=x'/z^2$:
$$y_0^2 = -7(x_0+1)(x_0^3+8x_0^2+5x_0-1)$$
Makedon't correspond to solutions of the change $x_0 \leftarrow (7\frac{x}{z^2}+7)^{-1}$original equation, then for somei.e. $y_0$:
$$y_0^2 = x_0^3+56x_0^2+245x_0-343$$$x=1$ or $x=7$).
This already appears above, and indeedTherefore, the original equation has no rational solutions.
ThisHere is much more tedious than I initially anticipated. I hope there are no more mistakes. Presumably, one might be able to replace the above "around-7" discussion by something more similar to the final discusion, i.e. only use reduction to solutions on another elliptic curve that has no solutions. If that does work out, I'll publish a tidy sage worksheet.code of the computation:
def cubic_to_ellipticcurve(f):
a, l = f.coeffs()[-1], f.coeffs()
if a != 1:
return EllipticCurve([0,l[2],0,l[1]*a,l[0]*a^2])
return EllipticCurve([0,l[2],0,l[1],l[0]])
def quartic_to_ellipticcurve(f):
for fac in factor(f):
if fac[0].degree() == 1:
r = fac[0].roots()[0][0]
v = f.variables()[0]
f = f(v+r)
f = sum([f.coeffs()[4-i]*v^i for i in [0..3]])
return cubic_to_ellipticcurve(f)
return None
R.<x> = QQ[]
possible_sels = []
for a in [-7,-1,1,7]:
for b in [-7,-1,1,7]:
c = a*b
E1 = cubic_to_ellipticcurve(c*(x^3+56*x^2+245*x-343))
if E1.rank() == 0 and E1.torsion_order() == 1:
continue
E2 = quartic_to_ellipticcurve(a*c*x*(x^3+56*x^2+245*x-343))
if E2.rank() == 0 and E2.torsion_order() == 1:
continue
E3 = quartic_to_ellipticcurve(b*c*(x+7)*(x^3+56*x^2+245*x-343))
if E3.rank() == 0 and E3.torsion_order() == 1:
continue
possible_sels += [(a,b,c)]
print possible_sels # prints []