Find all pairs of positive integers $a$ and $b$ so that for every prime p coprime to $b, a \bmod p < b \bmod p$? To clarify, $a\bmod p$ is defined to be the unique $r\in \{0,\cdots p-1\}$ so that $r\equiv a\bmod p$.
Edit: Initially, I thought it was too hard to find all solutions, so I only asked for infinitely many solutions. But now I see it's utterly trivial to find infinitely many, and because I'd like a better understanding of this problem, I'd like to ask for all solutions.
There is a nontrivial theorem that if $a$ and $b$ are positive integers such that $a\mod p \leq b\bmod p$ for all primes $p$, then $a = b$, and this problem was inspired by that theorem.
Call a pair $(a,b)$ of positive integers good if for every prime p coprime to b, $a\bmod p < b\mod p$. Clearly if there exists a prime so that $b\equiv 1\bmod p,$ then a must be divisible by that prime.
By choosing $p$ to be sufficiently large, we see that $a<b$. It suffices to check that $a\bmod p < b\bmod p$ for all primes coprime to b, it suffices to verify that $a\bmod p < b\bmod p$ for all primes less than b that are coprime to b. If a and b are both odd, then $a\mod 2 = b\bmod 2 = 1,$ contradicting the inequality. Hence at least one of $a$ and $b$ is even. $a$ cannot be congruent to $p-1\bmod p$ for any prime p coprime to $b$ as otherwise $a\bmod p\ge b\mod p.$ Suppose $a$ is even. Then a is at least $4.$ If $b=5,$ then $a\bmod k < b\bmod k$ for $k=2,3$. If b is even. We can choose $a=4,b=5$ for this to hold. Now if $b=6,$ we need $a\bmod l < 6\bmod k$ for $k=5$ and $a<6$. We can just choose $a=5.$ If $b=7,$ we need $a\bmod l < b\bmod k$ for $k=2,3,5$ and $a<b$. So $a$ is even and congruent to 0 mod 3.
Another method I was thinking of is to use the CRT to find solutions to a system of congruences $b\equiv x_i\bmod p_i$ for some distinct primes $p_i$, though I'm not sure how to make significant progress using this approach. The issue is that $b$ could be much larger than all the given $p_i$'s, and so there could be many primes less than b that would need to be checked.