need some help to prove that for each n, there are n consecutive integers, each of which is divisible by perfect square larger than 1.
Try induction? Only consider squares of primes. For $\displaystyle n=0,1,2$, the proof is almost trivial (with 8,9 divisible by 4,9 respectively). So, if you are looking for three numbers divisible by perfect squares larger than 1, you can look for $\displaystyle k, k+1, k+2$ where $\displaystyle k \equiv 0 \pmod{4}, k+1 \equiv 0 \pmod{9}, k+2 \equiv 0 \pmod{25}$. Now, you apply the Chinese Remainder Theorem: $\displaystyle k \equiv 548 \pmod{900}$. So, $\displaystyle 548,549,550$ are divisible by $\displaystyle 2^2,3^2,5^2$ respectively. So, for the $\displaystyle n$-th case, let $\displaystyle p_1,\ldots p_n$ be distinct primes, and solve the Chinese Remainder Theorem for $\displaystyle k \equiv 0\pmod{p_1^2}, \ldots, k+n-1 \equiv 0 \pmod{p_n^2}$. Since each prime you chose is distinct, the theorem guarantees that a solution will exist, so you are done.