# The rational points on the curve: $y^2=ax^4+bx^2+c$.

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 :$y^2=ax^3+bx^2+cx+d$?

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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 $y^2=ax^3+bx^2+cx+d$? –  Next 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. –  Next Mar 19 '13 at 12:34
@Gerry Myerson:I have known how to do it,thanks a lot! –  Next Mar 19 '13 at 13:12

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.

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Thank you,your answer is very clear to understand. –  Next Mar 19 '13 at 14:17

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.

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Thank you very much,I have got it with your help. –  Next Mar 19 '13 at 14:17

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?

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