If $a,b,c \in \mathbb{Z}$ are distinct, then there are infinitely many $n \in \mathbb{Z}$ such that $a+n$, $b+n$ and $c+n$ are pairwise relatively prime.
I'm not sure how to solve this problem. Here's what I did so far:
We note first that $$\begin{cases} \operatorname{gcd}(x+a, x+b) = \operatorname{gcd}(a-b, x+b)\\ \operatorname{gcd}(x+b, x+c) = \operatorname{gcd}(b-c, x+c)\\ \operatorname{gcd}(x+a, x+c) = \operatorname{gcd}(c-a, x+a) \end{cases}.$$ Let $a-b = p_1^{n_1}\ldots p_k^{n_k}$, $b-c = q_1^{m_1}\ldots q_s^{m_s}$ and $c-a = r_1^{l_1} \ldots r_t^{l_t}$ be the prime factorization of each difference. We can now set up the following system of congruences: $$\begin{cases} x+b \equiv 1 \mod p_j, j = 1, \ldots, k\\ x+a \equiv 1 \mod r_j, j = 1, \ldots, t\\ x+c \equiv 1 \mod q_j, j = 1, \ldots, s \end{cases}.$$ A solution to the system above would imply that $x+a$, $x+b$ and $x+c$ are pairwise relatively prime. Indeed, given that $x+b$ is not divisible by any prime in the prime factorization of $a-b$, we must have $\operatorname{gcd}(x+a, x+b) = 1$ by the initial observation. This argument also holds for the other sums. The only issue with the system above is that we may have indices $i, j$ such that $p_i = r_j$, for example, since they belong to a different prime factorization. Here, however, I thought about using the Generalized Chinese Remainder Theorem. It seems to be applicable because, for instance, if $p_i = r_j$, then $\operatorname{gcd}(p_i, r_j) = p_i = r_j$. Also, $(1-b) - (1-a) = a-b$, which means $\operatorname{gcd}(p_i, r_j)$ divides $(1-b) - (1-a)$. Since there is a solution by the GCRT, say $x_0$, there are infinitely many: $x = x_0 + n (p_1\ldots p_k q_1 \ldots q_s r_1 \ldots r_t), n \in \mathbb{Z}$ (of course, removing the duplicate primes in this product).
Does this work or am I missing something?
Please, don't close this question. I know this question has already been asked, but my attempt is significantly different from the others, so I'd like some feedback on what I did.
