How to solve arithmetic problems involving infinite sets of integers such as this one? Let $A\subset \mathbb{N}$ be an infinite set. Let $N = 10^{2020}$. Prove that :
$$ \exists(n,m)\in A^2,\quad \exists p \;\text{prime} \geq N,\quad p|n+m$$
I couldn't manage to solve this problem. How to do it?
I thought of supposing that $\forall(n,m)\in A^2,\quad \forall p \;\text{prime}\quad p|n+m \quad \Rightarrow \quad p <N$ and see where it ends but i'm not sure how to continue.
Could you help me?
Thanks
 A: I actually found a proof of this result! Here it goes:
As $A$ is a subset of $\mathbb{N}$, we can enumerate its elements: $A = \{a_1,a_2,\dots\}$ with $a_i<a_{i+1}$ for all $i$.
Let $A+A = \{a+b\,\mid\, a\in A, b\in A\}$, and for all set $E\subset\mathbb{N}$,
$P(E) = \{p\,\text{prime}\,\mid \exists x\in E,\, p|x\}$.
We want to show that $P(A+A)$ is infinite.
Let's assume that $P(A+A)$ is finite, of cardinality $k$. We then have $P(A+A) = \{p_1,p_2,\dots,p_k\}$.
Every prime $p$ is greater than $2$, so $p^{\alpha} \to +\infty$ as $\alpha \to +\infty$. As a result, we can choose $\alpha_1,\dots,\alpha_k\in \mathbb{N}$ such that
$$\forall i\in \{1,2,\dots, k\},\quad p_i^{\alpha_i}>a_{k+1}-a_1\quad \text{where $a_{k+1}$ is the $k+1$-th element of $A$}$$
Let $N = \prod_{i=1}^kp_i^{\alpha_i}$. As $A$ is infinite, we can choose $a\in A$ such that $a>N$.
Let $i\in \{1,\dots, k+1\}$. If $$\forall j\in \{1,\dots, k\},\, v_{p_j}(a+a_i)\leq\alpha_j$$ then $$\prod_{j=1}^k p_j^{v_{p_j}(a+a_i)} \leq \prod_{j=1}^k p_j^{\alpha_j} \quad \Leftrightarrow \quad a+a_i \leq N$$
which is absurd as $a+a_i\geq a>N$. Therefore, there exists $j_i\in \{1,\dots, k\}$ such that $v_{p_j}(a+a_i)>\alpha_j$.
With $\beta_{j_i} = v_{p_j}(a+a_i)$, we have:
$$\forall i\in \{1,2,\dots, k+1\},\quad \exists j_i\in \{1,\dots, k\},\quad \exists \beta_{j_i}>\alpha_j,\quad p_{j_i}^{\beta_{j_i}}|a+a_i$$
By pigeonhole principle, there exists two differents values of $i$ for which $j$ is the same (we set $i=r$ and $i=s$ be the two values such that $j_r = j_s := j$). Then $\beta_{j_r} = \beta_{j_s} = \beta_{j}$, so:
$$p_j^{\beta_j}|a+a_r\quad \text{and}\quad p_j^{\beta_j}|a+a_s\quad \text{(we remind the reader that $r\ne s$ so $a_r\ne a_s$)}$$
Because $\alpha_j<\beta_j$, we conclude that $p_j^{\alpha_j}$ divides both $a+a_r$ and $a+a_s$, so $p_j^{\alpha_j}|a_s-a_r$ therefore $p_j^{\alpha_j}\leq|a_s-a_r|$.
But we also know that $p_j^{\alpha_j} >a_{k+1}-a_1 \geq |a_s-a_r|>0$. This is absurd, so the assumption that $P(A+A)$ is finite is false.
Therefore $P(A+A)$ is infinite.
