# Prove that for each $n$, there exist $n$ consecutive integers, each of which is divisible by a perfect square larger than $1$.

Prove that for each $n$, there exist $n$ consecutive integers, each of which is divisible by a perfect square larger than $1$. Can any one help with this? I was stuck because I know how to solve for a particular $n$ value, but I was helpless in the generalized form.

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## 1 Answer

Use the Chinese Remainder Theorem on the congruences:

$$x \equiv 0 \pmod{2^2} \\ x+1 \equiv 0 \pmod{3^2} \\ x+2 \equiv 0 \pmod{5^2} \\ \dots$$

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