Q:

for every positive integer $n$, find $n$ consecutive numbers which are not squarefree

I want to show that for every positive integer $n$, I can find $n$ integers in a row whose prime factorization contains at least one prime twice.
I checked the cases $n=2$ $(8,9)$ and $n=3$ $(48,49,50)$ but I don't have an idea how to find a general proof.

A:

Let $p_0,p_1,\dots, p_{n-1}$  be distinct primes.  Consider the system of congruences
$$x\equiv -i \pmod{p_i^2}$$
($i=0,1,2,\dots, n-1$).
By the Chinese Remainder Theorem, this system has a solution $a$. Note that $a$ is divisible by $p_0^2$, $a+1$ is divisible by $p_1^2$,  and so on.

