Integer points in a curve?

The question asks for necessary and sufficient conditions for a given curve to have integer points, that is, $(x_k,y_k)$ such that $x_k,y_k\in\mathbb{Z}$. For example, a necessary and sufficient condition for the curve $\mathcal{C}\colon y=ax^2+bx+c$ to have infinitely many integer points is precisely that it contain at least three integer points. Fewer than three is not enough, since for example $y=\frac{2x^2+1}{2}$, $y=\sqrt{2}x^2$, and $y=\sqrt{2}(x^2-1)$ are counterexamples to the claim.

I've also done the linear case.

However I would like to know what conditions a given cubic, or even a conic, must satisfy, e.g., what are necessary and sufficient conditions for a curve $f(x)=ax^3+bx^2+cx+d$ with $a\neq 0$? How many integer points are required so that we know the curve has infinitely many integer points?

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I'm sorry if English is your second language, but the question isn't very clear. Could you include some diagrams or equations to help? –  Robert Mastragostino Jun 3 '12 at 2:03
Please, edit my question because I'm terrible with my english... –  Mario De León Urbina Jun 3 '12 at 2:04
Or you could post in your native language. Someone might translate. –  MJD Jun 3 '12 at 2:04
Don't explain in a comment. Edit the question. –  MJD Jun 3 '12 at 2:08
@Mario Escribe tu pregunta en español, yo trataré de editarla y escribirla en inglés. Procura incluir toda la información necesaria. –  DonAntonio Jun 3 '12 at 2:12

For $y=ax^3+bx^2+cx+d$, four is enough, three is not.

The counterexamples for 3 are much as in the quadratic case. Let $p(x)$ be a cubic with integer coefficients and three integer roots, and consider $\sqrt2p(x)$.

If there are 4 then each integer point gives you a linear equation in the 4 unknowns $a,b,c,d$, with all coefficients integers, so all the unknowns must be rational (by Cramer's Rule). Putting over a common denominator, $T$, we have $y=(Ax^3+Bx^2+Cx+D)/T$, with $A,B,C,D,T$ all integers. If this quotient is an integer for $x=k$, then it is an integer for $x=k+nT$ for all integers $n$ (just plug in $x=k+nT$, multiply everything out, and see what happens --- or think about it in terms of congruences modulo $T$).

The same arguments work for polynomials of any degree.

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It suffices to show that $\rm\:f\in \mathbb Q[x]\:$ since then $\rm\:f = g(x)/m,\:$ for $\rm\: g\in \mathbb Z[x],\ m\in\mathbb Z,\:$ therefore $$\rm f(n) = \frac{g(n)}m \in \mathbb Z\ \:\Rightarrow\:\ f(n\!+\!m\,\mathbb Z) = \frac{g(n\!+\!m\,\mathbb Z)}m\in \mathbb Z\quad by\quad g(n\!+\!m\,\mathbb Z)\equiv g(n)\:\ (mod\ m)$$

If $\rm\:f\:$ has $\rm\:1\!+\!deg\:f\:$ rational points $\rm (q_i,r_i)$ then $\rm\:f\!-\!r_1\:$ has root $\rm\:x = q_1,\:$ so $\rm\:f\!-\!r_1 = (x\!-\!q_1) g.\:$ $\rm g = \dfrac{f\!-\!r_1}{x\!-\!q_1}$ has $\rm1\!+\!deg\:g\:$ points $\rm (q_i,r_i)_{ i \ge 2}.$ Induction $\rm\Rightarrow g\in \mathbb Q[x]\Rightarrow f = r_1\! +\! (x\!-\!q_1)g \in \mathbb Q[x].$

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