Timeline for Which Diophantine equations can be solved using continued fractions?
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| Oct 19, 2011 at 11:02 | comment | added | Samuel Hambleton | If there are such polynomials $T(t, u)$ and $U(t, u)$, then points from the "parametrization" yield principal forms and principal ideals. I haven't found any such polynomials. | |
| Oct 19, 2011 at 10:57 | comment | added | Samuel Hambleton | Thanks. Are you saying that the binary quadratic form $3 x^2 + 13 x y -5 y^2$ comes from the parametrization? From Yamamoto's work, which appears in Buell's book, the binary quadratic form $(z, x, z^2)$ is discussed, where $x^2 - \Delta y^2 = 4 z^3$ and $gcd(x, z) = 1$. Franz Lemmermeyer and I looked at $z T^2 + x T U + z^2 U^2 = \Delta y^2$. If rel. prime integers $T, U$ exist satisfying this, then $(z, x, z^2)$ is principle. I think there should be polynomials $T(t,u)$ and $U(t, u)$ : $$((t^2 - Δ u^2)/4)T^2 + ((t^3 + 3 Δ t u^2)/4) T U + ((t^2 - Δ u^2)/4)^2U^2 = Δ ((3t^2 u + Δ u^3)/4)^2 ,$$ | |
| Oct 19, 2011 at 8:35 | history | edited | Will Jagy | CC BY-SA 3.0 |
added 91 characters in body
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| Oct 19, 2011 at 8:29 | history | answered | Will Jagy | CC BY-SA 3.0 |