Suppose a curve of degree $d$ in the plane passes through $\frac{d^2+3d}{2}$ lattice points. Must it pass through another lattice point?
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$\begingroup$ I think you mean "Must the curve pass through another point of $\mathbb Z^2$?". Is that correct? $\endgroup$– LSpiceCommented Jun 6, 2017 at 2:02
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$\begingroup$ yes, I think so. Thank to You very much @LSpice $\endgroup$– Bùi Thị LanCommented Jun 6, 2017 at 3:54
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1 Answer
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Yes if $d=1$, but no for $d\ge 2$.
Schinzel's Theorem says that for every $n\in \mathbb{N}$, there is a circle with exactly $n$ lattice points on the circumference. In particular, there is a circle passing through exactly $5$ lattice points, not $6$, which is a counterexample in degree $2$: $$\left(x-\frac{1}{3}\right)^2 +y^2 = \frac{5^4}{9}$$
Circles containing more points can be used to construct reducible counterexamples with higher degrees, as can curves like $xy=m$ where we can control the number of factors of $m$.
answered Jun 6, 2017 at 4:30
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$\begingroup$ Dear Dr. @DouglasZare, I want to ask You with modify question, could You answer me: Let $z=f(x,y) \in R[x,y]$, deg $f = n$, If there exists $k =\frac{n^2+3n}{2}$ integral roots $(x_i, y_i)$ (ingeneral position) of $z$ is (then) there exist another integral point $(u,v)$ such that $z = f(u,v) \in Z$ ? $\endgroup$ Commented Jun 6, 2017 at 12:07
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$\begingroup$ @Bùi Thị Lan: Exactly what do you mean by "in general position?" There are several possible definitions. $\endgroup$ Commented Jun 6, 2017 at 18:24
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$\begingroup$ If the counterexamples above count as having integer roots in general position by whatever definition you are using, then only countably many multiples would take a nonzero integer value at a lattice point, so any other multiple is a counterexample to the modified version. $\endgroup$ Commented Jun 6, 2017 at 18:49