Do these irrationals exist?

Question

An irrational $a$ verifies : $\{a\times n+k;(n,k)\in\mathbb Z^2 \}$ is dense in $\mathbb R$.

If you take $a$ universe then : $\forall b\in \mathbb N^*, \{a\times n^{b}+k;(n,k)\in\mathbb Z^2\}=A(a,b)$ is dense in $\mathbb R$.

1. Does exists $a$ irrational with : $\{a \times n^2+k ; (n,k) \in \mathbb Z^2 \}=A(a,2)$ isn't dense in $\mathbb R$ ?

2. Let $a \in \mathbb R -\mathbb Q$. Is it easy to find $C(a)=\{b\in\mathbb N^*,A(a,b) \text{ isn't dense in }\mathbb R$} ?

Answer 1

If I understand, your Question 1 asks "Does there exist an irrational number $\alpha$ such that $\{n^2\alpha +k :n, k \in \mathbb Z\}$ is not dense in $\mathbb R$?" The answer is no. This fact is usually deduced from a stronger condition on the sequence $(n^2\alpha)_{n\in \mathbb N}$, called uniform distribution mod $1$, cf. Wikipedia. When $(x_n)_{n\in \mathbb N}$ is uniformly distributed mod 1, the numbers $\{x_n+k: n\in \mathbb N, k\in \mathbb Z\}$ are dense in $\mathbb R$ (but the reverse implication can fail). The classic text of Kuipers and Niederreiter is the standard introduction - see Theorem 3.2 in Chapter 1. This may also answer your Question 2.

Kuipers, L.; Niederreiter, Harald, Uniform distribution of sequences, Pure and Applied Mathematics. New York etc.: John Wiley & Sons, a Wiley-Interscience Publication. xiv, 390 p. (1974). ZBL0281.10001.

Answer 2

I'm not sure I'm understanding your question (as for others, I'm confused about 'universe'), but for every irrational $a$ and positive integer $b$, your set $A(a,b)$ is dense. So $C(a)$ is always the empty set.

This seems to already be stated in your post, but maybe it's related to not knowing what 'universe' means.

EDIT: adding a few more details at request of the OP. The easiest self-contained proof uses the so-called van der Corput lemma, which states that if $(x_n)$ is a sequence in $[0,1)$ and for all $h \in \mathbb{N}$, the sequence $y^{(h)}_n = (x_{n+h} - x_n) \pmod 1$ is equidistributed, then $(x_n)$ itself is equidistributed.

Now you can easily prove the following by induction on the degree: for any non-trivial polynomial $p(n)$ with integer coefficients and any irrational $a$, the sequence $p(n) a \pmod 1$ is equidistributed. This obviously implies that $\{p(n) a + k \ : \ k,n \in \mathbb{Z}\}$ is dense in $\mathbb{R}$.

I'm sure there are many resources, but here is a nice expository paper on van der Corput and more: https://arxiv.org/pdf/1510.07332.pdf.