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 , 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 where . Now, you apply the Chinese Remainder Theorem: . So, are divisible by respectively. So, for the -th case, let be distinct primes, and solve the Chinese Remainder Theorem for . Since each prime you chose is distinct, the theorem guarantees that a solution will exist, so you are done.