Is there known a number field $K$ and a curve $F(x, y) \in K(t)[x, y]$ such that $F(x, y)$ does not have points over the field of rational functions $K(t)$ but for all but finitely many positive integer values of $t$ the respective specialization has a point over $K$?

(I expect the answer to be "such a $K$ is not known" and the problem of similar difficulty as that for $K = \mathbb{Q}$ but have not seen an expert comment on it so far.)

Thank you.


Assuming the parity conjecture (known to be a consequence of the BSD conjecture) for "congruent number curves" $E_d: d \cdot Y^2 = X^3 - X$, you can even take $K = {\bf Q}$ and $$ F(x,y) = 4x^2 + 1 + (8t-1) x y^2. $$

The curve $F(x,y) = 0$ is birational with the elliptic curve $(8t-1) Y^2 = X^3 + 4X$ (let $x=1/X$ and $y=Y/X$), which in turn is 2-isogenous with $E_{8t-1}: (8t-1) Y^2 = X^3 - X$. The only ${\bf Q}(t)$-rational points of the curve $(8t-1) Y^2 = X^3 + 4X$ are the torsion points at $(0,0)$ and infinity, and I chose coordinates $X,Y$ that put both of these points on the line at infinity, so there are no finite solutions of $F(x,y) = 0$. On the other hand, if $t$ is a positive integer then $|8t-1| \equiv 7 \bmod 8$, so $E_{8t-1}$ has sign $-1$, whence under BSD it has odd (and thus positive) rank over $\bf Q$, so infinitely many rational points.

[This example has no exceptional $t$, but it's clear a priori that if there's an example with only finitely many exceptional $t$ then one can translate $t$ to get a "new" $F$ that has none.]

  • $\begingroup$ Thank you very much for the answer. I was aware that the BSD conjecture implies a positive answer for $K = \mathbb{Q}$ and wanted only to find out whether the additional freedom of varying the $K$ isn't really expected to make the problem much more accessible. Also, thank you, David, for the promotion. $\endgroup$ – streetcar277 Jun 19 '14 at 11:40

The complexity arises because not solve such equations. Usually use of modular arithmetic. Although quite often it is not able to give the formula of the solution.

We give some examples.

Unique Integer solution of a non-linear equation - sometimes it is possible to write a simple formula.

How to find $a,b\in\mathbb{N}$ such that $c = \frac{(a+b)(a+b+1)}{2} + b$ for a given $c\in\mathbb{N}$ - or like this.

Diophantine equation: $(x-y)^2=x+y$ - or like this.

Perfect Square relationship with no solutions - for some special cases you can get the formula.

How to solve an equation of the form $ax^2 - by^2 + cx - dy + e =0$? - for some special cases you can get the formula.

How can I solve equation $x^2 - y^2 -2xy - x + y = 0$? - you can consider this simple.

Proving a Pellian connection in the divisibility condition $(a^2+b^2+1) \mid 2(2ab+1)$ - you can consider this simple.

Existence of $x,y$ Satisfying Diophantine Equation - you can consider this simple.

The system of genus characters determined by a binary quadratic form - there are written the same formula.

I think enough. There's formula for binary quadratic forms. Usually only for this form can write formulas of solutions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.