Must a polynomial function of $x$ pass through infinitely many integer lattice points?

Question

I made a mistake in my formulation of this question when I last asked it and got downvoted because the answer was actually trivial. However, I think the intended question is actually an interesting one, so I'm posting it again. Let $C$ be a planar curve defined by a polynomial function of $x$ with rational coefficients. Must $C$ pass through infinitely many integer lattice points? I'm assuming we know little about whether $C$ must pass through infinitely many Gaussian primes, but if you know anything about that I'd love to hear it.

Answer 1

Yes, if if $C$ is known to pass through at least one lattice point.

Let $$y=\sum \frac{a_k}{b_k}x^k$$ where $a_k\in \Bbb Z$, $b_k\in\Bbb N$. If this curve passes through $(0,0)$, let $m=\prod b_k$. Then $y\in \Bbb Z$ at least for $x\in m\Bbb Z$. If the curve passes trhogh another lattice point $(a,b)$, translating the curve to pass through $(0,0)$ instead still leaves us with a rational polynomial so that the preceeding argument applies.

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However, there are rational polynomials that totally miss all lattice points (I totally missed this point until Vik78 pointed me to that). Take for example $$ y=\frac 12+x^{42}.$$ Wheneever $x$ is an integer, $y$ is not.