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I wonder how to find the rational points on the curve: $y^2=ax^4+bx^2+c$.

Is there infinite rational points on this curve?

For example:$y^2=x^4+3x^2+1.$If we set $y=x^2+k$,then $2kx^2+k^2=3x^2+1$, Can one turn the equation to the form :$y^2=ax^3+bx^2+cx+d$?

Thanks in advance.

lsr314
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  • It is an elliptic curve problem. What transformation does it need? – Gerry Myerson Mar 19 '13 at 11:56
  • @Gerry Myerson:I wonder can we turn the equation to the form as $y^2=ax^3+bx^2+cx+d$? – lsr314 Mar 19 '13 at 12:00
  • Yes, there is a procedure for doing this. Unfortunately, I'm away from my references, and not up to doing it from memory. But...what advantage do you get by turning it into the form you want? – Gerry Myerson Mar 19 '13 at 12:28
  • @Gerry Myerson:To get more solutions from a given solution. – lsr314 Mar 19 '13 at 12:34
  • @Gerry Myerson:I have known how to do it,thanks a lot! – lsr314 Mar 19 '13 at 13:12
  • Good! Now why not share it with us, either for the special case $y^2=x^4+3x^2+1$, or for the general $y^2=ax^4+bx^2+c$? – Gerry Myerson Mar 19 '13 at 22:27
  • @Gerry Myerson:I'd like to do it with the example.$y^2=x^4+3x^2+1$,set $U=y-x^2-3/2,V=x(y-x^2-3/2)$,then $x=V/U,y=U+(V/U)^2+3/2$,rewrite our given equation $y^2-(x^2+3/2)^2=1-(3/2)^2=-5/4,(y+x^2+3/2)(y-x^2-3/2)=-5/4,U(U+2(V/U)^2+3)=-5/4,U^3+2V^2+3U^2=-5/4U,V^2=-1/2U^3-3/2U^2-5/8U.$ – lsr314 Mar 20 '13 at 05:18
  • @Isr314 , is there any reference to study thrse kind of transformation of Any problems to y^2 = Cubic in x. – Guruprasad Aug 29 '24 at 00:51

3 Answers3

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You can find some changes of variables to transform a quartic hyperelliptic curve into a Weierstrass equation at

  • Page 77 of: Mordell, Diophantine Equations, Academic Press, New York, 1969.

  • Page 37 of: L. Washington, Elliptic Curves: Number Theory and Cryptography (Discrete Mathematics and Its Applications), Chapman & Hall, 2003.

The results are quoted in my article with Scott Arms and Steven Miller, Appendix B, page 17.

Álvaro Lozano-Robledo
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You can turn $y^2 = a x^4 + b x^2 + c$ into $y^2 = x^3 + px + q$ assuming you can find one rational point on $y^2 = a x^4 + b x^2 +c$. The easiest case is when $a$ is square. I do an example of this computation here.

Community
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David E Speyer
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You can search for points on this sort of equation (of any degree) using Sage-s interface to Michael Stoll's ratpoints C program, but it is hidden:

sage: from sage.libs.ratpoints import ratpoints
sage: ratpoints([1,0,3,0,1],1000)
[(1, 1, 0), (1, -1, 0), (0, 1, 1), (0, -1, 1)]

Now this is the equation of a curve of genus 1, so if it has any rational points at all then it is isomorphic to its Jacobian which you can put into Weierstrass form using standard formulas. This can be done in Sage: try Jacobian?

John Cremona
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