# Are there any rational solutions to this equation?

I am not sure if this is an appropriate question, but I was asked this by a colleague today and do not know how to answer it.

1) Are there any rational solutions to the following equation: $$x^3-8x^2+5x+1 = -7y^2(x-1)x$$

2) Is it possible that this is an elliptic curve in disguise? I have noticed that after projectivizing, there are two points at infinity. Perhaps this is okay under some change of variables? (I plead ignorance on this.)

• sage claims that the genus is in fact 2. – Dror Speiser Mar 1 '11 at 21:29
• Changing $y$ to $\frac{y}{-7x(x-1)}$ and multiplying the resulting equation by $7x(x-1)$ gives a model $C: y^2=-7x(x-1)(x^3-8x^2+5x+1)$, of genus 2 (quintic with distinct roots), as Dror and William say. According to Magma, the Jacobian of $C$ has rank 2 and $C$ has points everywhere locally. I am not an expert, but I think this is considered to be the difficult case' (rk J$\ge$genus), so it could be hard to prove that there are no rational solutions. – Tim Dokchitser Mar 1 '11 at 22:28
• Does there exist a morphism $C \to E$ (in other words, is the Jacobian of $C$ a product of elliptic curves?). If so, then this provides a necessary condition for the existence of rational points on $C$. – François Brunault Mar 7 '11 at 2:09
• The study of rational points on hyperelliptic curves has been carried out by Bruin, Flynn, Stoll and others with some remarkable results. Your example is relatively simple--there is a covering set consisting of 2 rank 0 elliptic curves. In Trans. Amer. Math. Soc. 357 (2005) #11, 4329-4347, "Towers of 2-covers of hyperelliptic curves", Bruin and Flynn do a nasty example: yy=30x(x-1)(x+1)(xx-5x+1), showing that for the finite rational points x is 0,1,-1,5,6,1/2,1/3 or -1/8. – paul Monsky Mar 13 '11 at 13:01

[Complete revamp of answer. It is based on the one before, but is better!]

In Tim's hyperelliptic equation, make the change of variables $y$ to $y/(-7)^2$, and $x$ to $x/(-7)$, to get: $$y^2=x(x+7)(x^3+56x^2+245x-343)$$

For every prime $p$ with $v_p(x)<0$, the valuation must in fact be even, thus appears to an even power in the factorisation of each of the factors on the right (as non-zero rational numbers).

We need only consider the squarefree parts of the factors. Assume $p$ divides the numerator of at least two of the factors, to an odd power. By a small gcd calculation (3 in fact), we see that $p$ must be $7$.

Thus, we must have that each factor is a square times a number in $\{ -7, -1, 1, 7 \}$. For each triple $(a,b,c)$ of numbers in the set, with $abc$ a square, there is a possible element of the 2-Selmer group defined by the curve $C_{a,b,c}$ (in $\mathbb{P}^4$): $$x = au^2,$$ $$\\ x+7=bv^2,$$ $$x^3+56x^2+245x-343=cw^2$$

Some of these might not have points locally and will not define an element of the 2-Selmer. But we will not make any explicit local computations.

For each such triple, the curve $C_{a,b,c}$ has a morphism into each of the curves: $$C_{a,b,c}^1:\\ y^2=c(x^3+56x^2+245x-343)$$ $$C_{a,b,c}^2:\\ y^2=acx(x^3+56x^2+245x-343)$$ $$C_{a,b,c}^3:\\ y^2=bc(x+7)(x^3+56x^2+245x-343)$$

A sage computation shows that for each such triple $(a,b,c)$, at least one of these three curves has no rational points (other than ones at infinity, or points that don't correspond to solutions of the original equation, i.e. $x=1$ or $x=7$).

Therefore, the original equation has no rational solutions.

Here is the sage code of the computation:

def cubic_to_ellipticcurve(f):
a, l = f.coeffs()[-1], f.coeffs()
return EllipticCurve([0,l,0,l*a,l*a^2])

def quartic_to_ellipticcurve(f):
for fac in factor(f):
if fac.degree() == 1:
r = fac.roots()
v = f.variables()
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 []
`
• I think you've fallen into the same errors that I made. You write: "Hence x^3+...-343z^6 is either a square or 7(square)". But it also could be -(square) or -7(square). Also later you say "Assume the latter, so that x' is a square". But x' could be -(square). – paul Monsky Mar 6 '11 at 14:41
• Dror, I apologize for "accepting" and then "unaccepting" this answer. I want to see if someone can handle the remaining case. – Micah Milinovich Mar 6 '11 at 21:55
• @Micah: Legitimate. I think I have solved the remaining case, but will wait to see if any holes are revealed. – Dror Speiser Mar 7 '11 at 2:06
• @Dror: This revised answer looks good to me, and should be accepted. If you don't object, I'll edit my answer in the next few days to a simpler (in the sense of reduced reliance on Sage) version of your revised answer--one just has to look up 2 elliptic curves, one of conductor 392 and one of conductor 784 in Cremona's tables; everything else can be done by hand. – paul Monsky Mar 7 '11 at 13:07
• @Paul: Cool. It will be good to have a well written computation. – Dror Speiser Mar 7 '11 at 15:41

As Dror says, according to Sage it's a genus 2 curve. Homogenizing your equation gives a degree 4 projective plane curve $C$, with arithmetic genus thus $(d-1)(d-2)/2 = 3\cdot 2/2 = 3$. The geometric genus is the arithmetic genus minus a contribution for each singular point. There is exactly one singular point on this projective curve, and assuming it is simple we get genus 2. Drawing a picture of this singular point, indeed it looks simple. So the assertion that the projective genus is 2, which is output from Sage, is reasonable. If indeed the genus is 2, then this implies that your equation has only finitely many rational solutions (by Faltings's theorem).

There is no known algorithm to find all rational points on a curve of genus $\geq 2$ in general. But one can always look. So I wrote a little compiled Cython program to search for rational points $(x,y,z)$ on the homogenous equation, with $-B\leq x,y,z < B$ and $z>0$, and in 13 seconds found no points with $B=1000$.

• I don't believe it... shouldn't you have used C.rational_points(bound=1000) ?! This should really be a simple exercise in either 7-adics or the number field generated by the polynomial on the left. Sage says the polynomial generates the unique abelian extension of discriminant 7^2 and that the class number is 1. – Dror Speiser Mar 1 '11 at 22:05
• @Dror: Yes, C.rational_points(bound=1000) would give the same result; however it is stupid generic Python code and is literally almost 100,000 times slower than the little Cython function I wrote, so it would take 2 weeks instead of 13 seconds. I don't know what you mean by "This should really be a simple exercise in either 7-adics or the number field generated by the polynomial on the left..." The curve does have rational points (the two at infinity), so I don't see how to rule out rational points by working p-adically. Maybe you can rule out integral points, but that is different. – William Stein Mar 1 '11 at 22:23

I think Dror's nice argument can be made more transparent as follows. Suppose that $y^2= x(x+7)(x^3+56x^2+245x-343)$ with $x,y$ in $\bf Q$ and $x$ not $0$ or $-7$. One sees easily that at a prime $p$ other than $7$ neither $x$ nor $x+7$ can have odd ord, either negative or positive. It follows that $x^3+56x^2+245x-343$ is either a square or $7$(square) in $\bf Q$.

As Dror says, the first case can't occur. Suppose we're in the second. Write:

$x=7a$, $x^3+56x^2+245x-343=343b^2$, so that $x(x+7)=343c^2$ for some $c$ in $\bf Q$. Then:

$a$ is not $0$ or $-1$, $a(a+1)=7c^2$, and $b^2=a^3+8a^2+5a-1$. The second equation shows that $a$ is in the $7$-adic integers and is $0$ or $-1\pmod7$. Furthermore $a$ is a square or $7$(square) in the $7$-adic integers. So $a$ can only be $0 \pmod 7$. But then $b^2 = -1 \pmod 7$, which can't happen.

EDIT: This is quite wrong. But I think Dror made the same errors; see the comment I attached to his answer.

FURTHER EDIT: I'll write out a less computer-dependent version of Dror's excellent answer. The curve Y^2=X^3-8(X^2)+5X+1 has good reduction at each prime other than 2 or 7, and cuspidal reduction at 7. So its conductor is 49(power of 2), and the same is true of the twists:

A: 7(Y^2)=X^3-8(X^2)+5X+1

B:-7(Y^2)=X^3-8(X^2)+5X+1

(1)... The transformation X-->X+3 takes A to 7(Y^2)=X^3+X^2-16X-29 which is isomorphic to Y^2=X^3+7(X^2)-794X-9947. X-->X-2 takes this to Y^2=X^3+X^2-800X-8359. So A has the minimal Weierstrass model (0,1,0,-800,-8359), and B has the minimal model (0,-1,0,-800,8359). Looking up Cremona's tables for curves of conductor 49(a power of 2) we find that A is isomorphic to 392b, that B is isomorphic to 784d, and neither has any finite rational points.

(2a)... Consider the curve 7(Y^2)=X(X^3-8(X^2)+5X+1). The transformation X-->1/X,Y-->Y/X^2 takes this to 7(Y^2)=X^3+5(X^2)-8X+1. X-->X+1 takes this in turn to 7(Y^2)=X^3+8(X^2) +5X-1. Then X-->-X gives B. So the only finite rational point on our curve is (0,0).

(2b)... Next consider -7(Y^2)=(X-1)(X^3-8(X^2)+5X+1). X-->X+1 takes this to -7(Y^2)= X(X^3-5(X^2)-8X-1), and X-->-X takes this in turn to -7(Y^2)=X^4+5(X^3)-8(X^2)+X. Applying X-->1/X, Y-->Y/X^2 we get B again. So (1,0) is the only finite rational point on our curve.

THE PROOF: The conclusions of 1,2a and 2b give Speiser's result: there are no rational points (x,y) on Y^2=-7(X)(X-1)(X^3-8(X^2)+5X+1) other than (0,0) and (1,0). For let (x,y) be such a point. Then at any odd prime other than 7 the ord of x cannot be an odd number, either positive or negative, and the same holds for the ord of x-1. So each of x, x-1 and x^3-8(x^2)+5x+1 has one of the forms: square, -(square), 7(square), or -7(square). The conclusion of 1 above tells us that x^3-8(x^2)+5x+1 can only be a square or -(a square).

Suppose that x^3-8(x^2)+5x+1 is a square. Then x(x-1) is -7(square). So x is integral at p=7 and congruent to 0 or 1 mod 7 in the 7-adics. If x were congruent to 1, x^3-8(x^2)+5x+1 would not be a square. So x is congruent to 0, and x-1, being congruent to-1, can only be -(square). So x=7(square), and the same holds for x(x^3-8(x^2)+5x+1) contradicting the conclusion of 2a.

Suppose finally that x^3-8(x^2)+5x+1 is -(square). Then x(x-1)=7(square). So x is integral at p=7 and congruent to 0 or 1 mod 7 in the 7-adics. If x were congruent to 0, x^3-8(x^2)+5x+1 would not be the negative of a square. So x is congruent to 1, and can only be a square. Then x-1 is 7(square), and (x-1)(x^3-8(x^2)+5x+1) is -7(square). This contradicts the conclusion of 2b.

• paul, this would look so much nicer if you would just enclose all formulas in dollar signs. Compare -7(Y^2)=X(X^3-8(X^2)+5X+1) and $-7Y^2=X(X^3-8X^2+5X+1)$. The only difference is I've put a dollar sign at either end (OK, I also discarded some parentheses). – Gerry Myerson Mar 7 '11 at 22:41
• Gerry: I agree, but I've never used anything like laTex, and I have no idea how things would come out if I started throwing dollar signs around. Apologies. – paul Monsky Mar 8 '11 at 0:06