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

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closed as off-topic by Marc van Leeuwen, tetori, T. Bongers, azimut, Cameron Buie Oct 23 '13 at 6:32

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This kind of stuff is obvious from the CRT. Why should I want to prove it? –  Marc van Leeuwen Oct 23 '13 at 4:33
    
@Marcvanleeuwen you do not simply argue with the commandments of our lord –  chubakueno Oct 23 '13 at 4:58

1 Answer 1

Consider the system of congruences $x\equiv 0\pmod{2^2}$, $x\equiv -1\pmod{3^2}$, $x\equiv -2\pmod{5^2}$, and so on up to $x\equiv -(n-1)\pmod{p_{n-1}^2}$. (Here $p_0,p_1,\dots,p_{n-1}$ are the first $n$ primes.)

This system has a solution $x$, by the Chinese Remainder Theorem. Any such $x$ is divisible by $2^2$, and $x+1$ is divisible by $3^2$, and so on.

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