# Rational points or a Weierstrass model for degree 8 elliptic curve

Related to rationally derived polynomials.

Neither Maple nor Magma online couldn't solve it.

Choose $s,h \in \mathbb{Q}$.

I am looking for rational points (possibly of finite order) on this elliptic curve or a Weierstrass model for it:

4*u^8 + 4*u^7*v + 16*u^6*v^2 + 28*u^5*v^3 + 40*u^4*v^4 + 28*u^3*v^5 + 16*u^2*v^6 + 4*u*v^7 + 4*v^8 - 18*u^7*h - 30*u^6*v*h - 90*u^5*v^2*h - 150*u^4*v^3*h - 150*u^3*v^4*h - 90*u^2*v^5*h - 30*u*v^6*h - 18*v^7*h + 35*u^6*h^2 + 78*u^5*v*h^2 + 189*u^4*v^2*h^2 + 260*u^3*v^3*h^2 + 189*u^2*v^4*h^2 + 78*u*v^5*h^2 + 35*v^6*h^2 - 36*u^5*h^3 - 84*u^4*v*h^3 - 168*u^3*v^2*h^3 - 168*u^2*v^3*h^3 - 84*u*v^4*h^3 - 36*v^5*h^3 + 16*u^4*h^4 + 32*u^3*v*h^4 + 48*u^2*v^2*h^4 + 32*u*v^3*h^4 + 16*v^4*h^4 - 18*u^7*s - 30*u^6*v*s - 90*u^5*v^2*s - 150*u^4*v^3*s - 150*u^3*v^4*s - 90*u^2*v^5*s - 30*u*v^6*s - 18*v^7*s + 86*u^6*h*s + 204*u^5*v*h*s + 522*u^4*v^2*h*s + 680*u^3*v^3*h*s + 522*u^2*v^4*h*s + 204*u*v^5*h*s + 86*v^6*h*s - 180*u^5*h^2*s - 516*u^4*v*h^2*s - 1032*u^3*v^2*h^2*s - 1032*u^2*v^3*h^2*s - 516*u*v^4*h^2*s - 180*v^5*h^2*s + 200*u^4*h^3*s + 544*u^3*v*h^3*s + 816*u^2*v^2*h^3*s + 544*u*v^3*h^3*s + 200*v^4*h^3*s - 96*u^3*h^4*s - 192*u^2*v*h^4*s - 192*u*v^2*h^4*s - 96*v^3*h^4*s + 35*u^6*s^2 + 78*u^5*v*s^2 + 189*u^4*v^2*s^2 + 260*u^3*v^3*s^2 + 189*u^2*v^4*s^2 + 78*u*v^5*s^2 + 35*v^6*s^2 - 180*u^5*h*s^2 - 516*u^4*v*h*s^2 - 1032*u^3*v^2*h*s^2 - 1032*u^2*v^3*h*s^2 - 516*u*v^4*h*s^2 - 180*v^5*h*s^2 + 408*u^4*h^2*s^2 + 1248*u^3*v*h^2*s^2 + 1872*u^2*v^2*h^2*s^2 + 1248*u*v^3*h^2*s^2 + 408*v^4*h^2*s^2 - 480*u^3*h^3*s^2 - 1248*u^2*v*h^3*s^2 - 1248*u*v^2*h^3*s^2 - 480*v^3*h^3*s^2 + 240*u^2*h^4*s^2 + 384*u*v*h^4*s^2 + 240*v^2*h^4*s^2 - 36*u^5*s^3 - 84*u^4*v*s^3 - 168*u^3*v^2*s^3 - 168*u^2*v^3*s^3 - 84*u*v^4*s^3 - 36*v^5*s^3 + 200*u^4*h*s^3 + 544*u^3*v*h*s^3 + 816*u^2*v^2*h*s^3 + 544*u*v^3*h*s^3 + 200*v^4*h*s^3 - 480*u^3*h^2*s^3 - 1248*u^2*v*h^2*s^3 - 1248*u*v^2*h^2*s^3 - 480*v^3*h^2*s^3 + 576*u^2*h^3*s^3 + 1152*u*v*h^3*s^3 + 576*v^2*h^3*s^3 - 288*u*h^4*s^3 - 288*v*h^4*s^3 + 16*u^4*s^4 + 32*u^3*v*s^4 + 48*u^2*v^2*s^4 + 32*u*v^3*s^4 + 16*v^4*s^4 - 96*u^3*h*s^4 - 192*u^2*v*h*s^4 - 192*u*v^2*h*s^4 - 96*v^3*h*s^4 + 240*u^2*h^2*s^4 + 384*u*v*h^2*s^4 + 240*v^2*h^2*s^4 - 288*u*h^3*s^4 - 288*v*h^3*s^4 + 144*h^4*s^4


subject to the constraint $u,v,h$ are all distinct.

Some points are $(u,v,h,s)=(-t,-t,-t,t)$ or $(t,t,-t,t)$.

Modulo my errors it is conjectured that there are no points subject to the constraint.

Suspect that other points (if any) will be of large height.

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It might help if you explain where this equation comes from. This doesn't look like an elliptic curve but some sort of parametrization where fibers are curves. Maybe take some fibers and see how they work? Also, assuming Birch and Swinnerton-Dyer you can do calculations over finite fields to help create bounds on rank. –  Watson Ladd Oct 14 '13 at 15:45
@WatsonLadd I believe it is genus $1$ in general, for the other variables too. –  joro Oct 14 '13 at 17:19

If you substitute $v = v + u$, $h = h + u$ and $s = s + u$, the equation becomes much simpler and the variable $u$ goes away entirely! Then substituting $h = hv$ reduces the degree of $v$ down to $4$ (removing a factor of $v^4$) and following it up by $v = st$ gets rid of $s$ after removing a factor of $s^4$. So you end up with the curve $$(16t^4 - 96t^3 + 240t^2 - 288t + 144)h^4 + (-36t^4 + 200t^3 - 480t^2 + 576t - 288)h^3 + (35t^4 - 180t^3 + 408t^2 - 480t + 240)h^2 + (-18t^4 + 86t^3 - 180t^2 + 200t - 96)h + (4t^4 - 18t^3 + 35t^2 - 36t + 16) = 0$$ which has genus $1$. Perhaps you can do something from here.

Continued: following Mike Zieve's suggestion, let's show there are no real points. First, substituting in succession $t = t + 1$, $h= (h+1)/2$ and $t = (1+g)/(1-g)$ converts it into the nice symmetric (!) expression $$(9g^4+6g^2+1)h^4-8gh^3+(6g^4+12g^2+2)h^2+(-8g^3-8g)h+g^4+2g^2+1$$ which we want to show is positive everywhere. In other words, we need to show $$(3g^2 + 1)^2h^4 + (6g^4+12g^2+2)h^2 + (g^2 + 1)^2 \geq 8gh^3 + 8(g^2+1)gh.$$ We have $$(6g^4+12g^2+2) = 4/3 + 2/3 + 6g^4 + 12g^2 \geq 4/3 + 4g^2+ 12g^2$$ by AM-GM. Split up the LHS as $(3g^2 + 1)^2h^4 + 4/3h^2$ plus $16g^2h^2 + (g^2 + 1)^2$. The first expression is $$\geq 2(3g^2 + 1)|h|^3 \sqrt{4/3} >= 2\cdot2 \cdot \sqrt{3} |g| |h|^3 \cdot 2/\sqrt{3} \geq 8gh^3$$ and the second is $$\geq 8gh(g^2 + 1),$$ finishing the proof. I guess you still have to check that equality is never attained for rationals, and also the boundary conditions (stuff we divided by), but that is pretty straightforward.

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Does this have any real points? The left side was positive for the first several pairs $(h,t)$ I tried. –  Michael Zieve Oct 15 '13 at 6:53
Indeed, the left hand side vanishes in the two points $(h,t)=(1/2+\alpha/6, 3+\alpha)$ with $\alpha=\pm\sqrt{3}$, and seems to be positive elsewhere. –  Peter Mueller Oct 15 '13 at 14:17
Thanks. Are the known points the only ones your transformations lose? (The original curve has points). –  joro Oct 16 '13 at 5:01
@joro: I think you should work that out on your own -- and accept the answer once you checked if it is correct. –  Peter Mueller Oct 16 '13 at 14:44

A variant of Abhinav Kumar's solution again uses his reduction to looking at $$(9g^4+6g^2+1)h^4-8gh^3+(6g^4+12g^2+2)h^2+(-8g^3-8g)h+g^4+2g^2+1,$$ which equals $$\tag{1} \left(3g^2h^2 - g^2 + 4gh - h^2 - 1\right)^2 + 12\left(gh(g-h)\right)^2.$$ So this shows non-negativity. Furthermore, if (1) vanishes, then $gh(g-h)=0$. But the cases $g=0$, $h=0$, and $g=h$ yield $h^2+1=0$, $g^2+1=0$, and $(g^2+1)(2g^2-1)=0$, respectively. None of these has a rational solution.

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