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Math Help - Sequences of Consecutive Non-Squarefree Integers?

  1. #1
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    Sequences of Consecutive Non-Squarefree Integers?

    Hi, I am new to this forum and unfortunately I'm in a number theory class that is kicking my ass. I need some help anywhere I can get it. My homework problem is this:

    Prove that for any positive integer k, there exists a sequence of k consecutive positive non-squarefree (squareful?) integers.

    So I found this one algorithm, which is kind of similar to the Chinese Remainder Theorem, that says let a1=1, a2=3, a3=5, and list the primes like that. Pick a k.
    Then let n1=(a1)^2
    Add (a1)^2 repeatedly until you get a number n2 such that
    (n2)+1 is congruent to 0 (mod (a2)^2)
    Then to this number, add the product ((a1)^2)((a2)^2) repeatedly until you get n3 such that
    (n3)+2 is congruent to 0 (mod(a3)^2)
    and so on, until you reach nk such that
    (nk)+(k-1) is congruent to 0 (mod (ak)^2)

    Now, one thing I don't understand is why the numbers in between nk and (nk)+(k-1) are squareful. I know that they are because I've played around with this on the calculator, but I can't show why. For example, I've gotten up to k=4: 14749, 14750, 14751, and 14752. I can see why 14749 and 14752 are squareful, but I don't know why 14750 and 14751 are squareful.

    Also, if there is another solution out there, that would be great too!!! Thank you!
    Last edited by Chelsea17; March 1st 2011 at 05:56 PM.
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    You probably know that 1 is not prime, right?
    I didn't really pay attention to the algorithm, but you're right in saying it's the Chinese Remainder Theorem. Just solve the system

    x\equiv 0 \mod 2^2
    x+1\equiv 0 \mod 3^2
    ...
    x+n\equiv 0 \mod p_{n+1}^2
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