It seems to me that for any integer $n>0$, there exists an integer $x$ such that for all $n \ge i > 0$, $x+i$ is not squarefree.
Here's my argument:
- Let $x_0 = 7$, $m_0 = 72$
- For any integer $a$, $8+72a$ and $9+72a$ will not be squarefree.
- $10+72a$ is never divisible by $4$ or $9$ and $10+72a, 10+72(a+1), 10+72(a+2), \dots, 10+72(a+p)$ forms a complete residue system modulo $p$ when $p \ne 2,3$
For example $10+72(20) \equiv 0 \pmod {25}$
- Let $x_1 = 1447$, $m_1 = 1800$
- Now, for any integer $a, 1448+1800a, 1449+1800a, 1450+1800a$ are not squarefree.
- $1451+1880a$ is never divisible by $4$ or $9$ or $25$ and $1451+1800a, 1451+1800(a+1), \dots, 1451+1800p$ forms a complete residue system modulo $p$ when $p \ne 2,3,5$
We can repeat this process for any $n$ that we want so long as $x$ can be sufficiently large.
Is my reasoning correct?
An alternative argument would run as follows:
By the Chinese Remainder Theorem, there exists a positive integer $x$ such that $x \equiv -i \mod p_i^2$ for $1 \le i \le n$, where $p_1$, $\ldots$, $p_n$ are the first $n$ primes. It follows that none of $x+1$, $x+2$, $\ldots$, $x+n$ are squarefree.