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Are there any solutions to the equation $s^{2}(1+t^{2})^{2}+t^{2}(1+s^{2})^{2}=u^2$ where $s,t,u\in \mathbb{Q}$ and $0 < s,t<1$? If so, is there a simple way to parametrize them all?

If I am understanding the geometry behind this problem, even if we pick a specific value for $t$ we are left with an elliptic curve, and it is possible there could be infinitely many solutions.

This question arises from some related (and somewhat esoteric) questions I have about rational points on unit circles. But this question looked "pretty" enough, I thought I'd ask the experts here.

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I don't know about parametrising them all, but if I got my calculations right then setting t=1/2 gives an elliptic curve of rank 1 and I see plenty of points. For example (all with t=1/2) s=15/16 or 14911/369120. –  Kevin Buzzard May 16 '11 at 23:11
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Random specialisations of $t$ all seem to have rank at least 1, so probably the curve in $s$ and $u$ has rank 1 over $\mathbf{Q}(t)$ which means there will be, amongst other solutions, infinitely many parametric solutions in $s$ and $t$. –  Kevin Buzzard May 16 '11 at 23:24
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As Kevin Buzzard indicated, now the $(s,u)$ curve is isomorphic to $$y^2=x^3+(-2t^4-8t^2-2)x^2+(t^8+8t^6+14t^4+8t^2+1)x,$$ and a point on this is $(4t^2,2t^5-2t)$, not being torsion. Mapping back to your $(s,u)$, this is $$s={t^4-1\over 4t^2},u={t^8+8t^6-2t^4+8t^2+1\over 16t^3}.$$ There is also $$s={t-1\over t+1},u={(t^2+1)^2\over(t+1)^2},$$ deriving from $(x,y)=(t^4+2t^3+2t^2+2t+1,2t^5-2t)$. Further, Modulo primes like 11 and 19, along with the torsion results $x=0,(t^2+1)^2,t^4+6t^2+1$, these are the only $F_p$ solutions with degree bounded by 4, and so are the only rational ones there. –  Junkie May 17 '11 at 4:28
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There is a short GP package (based on a paper done by Mike Artin, Fernando Rodriguez-Villegas and John Tate) to pass from y^2=quartic to y^2=cubic (although some of this was done in an old paper by Hermite much before, I can post the reference if needed). Check the link ma.utexas.edu/cnt/cnt-frames.html (the file jacobians) –  A. Pacetti May 17 '11 at 23:53
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If $u^2=quartic(s)$ and we know a point, here $(0,t)$, you can follow Cassels 8(iii) in Lectures on Elliptic Curves, or computer algebra systems will do it, to get the rational isomorphism. I suspect we can show that the Mordell-Weil group over $Q(t)$ is only $(Z/2)^2\oplus Z$ by descent or cohomology, but the details maybe tricky, especially if by hand. Another approach, is noting the rank 1 specializations and use Silverman's theorem, III.11 in his Advanced Topics in the Arithmetic of Elliptic Curves, effectively. Finally, Rank increase upon specialization is difficult, generally. –  Junkie May 18 '11 at 2:07

1 Answer 1

up vote 6 down vote accepted

(Edited with more details.)

I know this is a really old question, but this has a nice connection to a problem considered by Euler, what is now known as Euler bricks, and I couldn't resist. The OP's equation is equivalent to finding three rationals $a,b,c$ such that,

$$\begin{aligned} a^2+b^2\; &= u_1^2\\ a^2-c^2\; &= u_2^2\\ b^2-c^2\; &= u_3^2 \end{aligned}\tag1$$

A solution (essentially by Euler) is,

$$a = \frac{s^2+1}{2s},\quad b = \frac{t^2+1}{2t},\quad c = 1\tag2$$

where $s,t$ must satisfy,

$$s^2(t^2+1)^2+t^2(s^2+1)^2 = w^2\tag3$$

and is the equation considered by the OP. For any solution to $(1)$ with $c\neq0$, then $s,t$ can be recovered as,

$$s = \frac{a\pm\sqrt{a^2-c^2}}{c} = \frac{a\pm u_2}{c}$$

$$t = \frac{b\pm\sqrt{b^2-c^2}}{c} = \frac{b\pm u_3}{c}$$

A small solution to $(3)$ is,

$$s = \frac{4p}{p^2-1},\quad t = \frac{3p^2+1}{p(p^2+3)}\tag4$$

This is the essentially the same one given by Euler for Euler bricks,

$$\begin{aligned} \alpha^2+\beta^2\; &= v_1^2\\ \alpha^2+\gamma^2\; &= v_2^2\\ \beta^2+\gamma^2\; &= v_3^2 \end{aligned}$$

where we have used $p \to p\sqrt{-1}$, and have correspondingly tweaked $(1)-(4)$. From this initial rational point $(4)$, one can then generate an infinite more.

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