Pitagar wrote: Two infinite arithmetic sequences with positive integers are given:$$a_1<a_2<a_3<\cdots ; b_1<b_2<b_3<\cdots$$It is known that there are infinitely many pairs of positive integers $(i,j)$ for which $i\leq j\leq i+2021$ and $a_i$ divides $b_j$. Prove that for every positive integer $i$ there exists a positive integer $j$ such that $a_i$ divides $b_j$. There is something wrong here. Let $a_i$ be $2,3,5,7,\dots$ and $b_i$ be $1,3^2,5^2,7^2,\dots.$ Clearly $a_i\mid b_i,\forall i\ge 2$ but $a_1$ does not divide any $b_i,i=1,2,\dots.$

@above they are arithmetic sequences. I didn't see that too and it also made me confused for a while.

You mean arithmetic progressions? Ok, it makes sense, they shouldn't increase too fast, otherwise a counterexample is possible, even for infinetely many $i.$

Let $r, s$ be the consecutive differences of the arithmetic progressions $a_n, b_n$, respectively. Lemma. $s/r$ is an integer. Proof. Let $c_n^{(k)} = b_{n+k}/a_n$, with $0 \le k \le 2021$. The "infinitely many pairs" condition implies that there are infinitely many $(n, k)$ so that $c_n^{(k)}$ is an integer. Therefore, since there are finitely many choices for $k$, there is a fixed $k$ so that there are infinitely many $n$ so that $c_n^{(k)}$ is an integer. However, for this fixed $k$, \[\lim_{n\to\infty} c_n^{(k)} = \frac{b_{n+k}}{a_n} = \frac{s}{r}.\]Therefore, $s/r$ must be an integer. With the same fixed $k$ as above, there exists some large $N$ so that $c^{(k)}_N$ is an integer, and since it is close to $s/r$, it must be equal to $s/r$. So, \[c^{(k)}_N = \frac{b_{N+k}}{a_N} = \frac{s}{r} = \frac{b_{N+k} + st}{a_N + rt} = \frac{b_{N+t+k}}{a_{N+t}} = c^{(k)}_{N+t},\]for any integer $t$. Therefore, $c_n^{(k)} = s/r$ for all integers $n$. Therefore, $a_i \mid b_{i+k}$ for all $i$.

Obviously there is a $k$ that is difference for infinitely many pairs $(i,j)$. Then $a_i \mid b_{i+k}$ for infinitely many $i$. The key fact is to note that $i-1=(a_i-a_1)/d_a=(b_{i+k}-b_1-k.d_b)/d_b$, and then $(a_i-a_1).d_b=(b_{i+k}-b_1-k.d_b).d_a$ (*). Using $a_i \mid b_{i+k}$ and the fact that those $i$ are infinitely many, $a_1.d_b-(b_1+k.d_b).d_a=0$ (**). So $a_i.d_b=b_{i+k}.d_a$ (1) by (*), so $d_a \mid d_b$. So $d_b=l.d_a$ (2) and $b_{i+k}/a_i=l$. So $l \mid b_1$ by (*). Thus $b_1=l.s$ (3), and by (1),(2),(3)(**), it's easy to calculate that $b_{i+k}=a_i.l$ for each $i$ (using that $a_1=s+k.d_a$).

I commented on it on my blog.

Denote the progressions by $an+b$ and $cn+d$. Consider the differences $j-i$ - they are infinitely many and the value of each is among $2022$ possible integers - hence there is an integer $0\leq k \leq 2021$ which equals infinitely many of these differences. In other words, for this $k$ we have $ai+b \mid c(i+k) + d$ for infinitely many $i$. We will now prove that this divisibility actually holds for all $i$. We necessarily have $ai+b \mid c(i+k) + d \mid ac(i+k) + ad$, hence $ai+b \mid ac(i+k) + ad - c(ai+b)$, i.e. $ai+b \mid ack + ad - bc$. As $a$, $b$, $c$, $d$ and $k$ are fixed while $i$ varies with infinitely many possible values, we must have $ack + ad - bc = 0$. Thus $c(i+k) + d = ci + \frac{bc}{a} = \frac{c}{a}(ai+b)$ and so $ai+b$ divides $c(i+k)+d$ if and only if $a$ divides $c$. Since this holds for at least one $i$, we get that $a$ divides $c$ and therefore the desired divisibility must hold for all $i$.

Lemma: for integers $a,b,c,d (a, c \ne 0)$, if $an+b\mid cn+d$ has infinitely many solutions $n\in \mathbb{N}$, then $an+b\mid cn+d$ for every $n\in \mathbb{N}$. Proof: if $n$ satisfies $an+b\mid cn+d$, then $$an+b \mid a(cn+d)-c(an+b) \implies an+b \mid ad-bc.$$if $ad-bc\ne 0$, then $|an+b|$ is bounded, hence finite solutions of $n$. Hence $ad=bc$, hence $\frac{cn+d}{an+b}=\frac{c}{a}$ is a constant. Hence it's always an integer. Lemma proven $\blacksquare$ Take $0\leqslant k\leqslant 2021$ such that $a_i\mid b_{i+k}$ has infinitely many solutions in $i\in \mathbb{N}$. By our lemma, we have $$a_i\mid b_{i+k} \text{ for all } i\in \mathbb{N}.$$Taking $j=i+k$, problem proven.