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Math Help - number theory

  1. #1
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    number theory

    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.
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  2. #2
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    Re: number theory

    Try induction? Only consider squares of primes. For 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 k, k+1, k+2 where k \equiv 0 \pmod{4}, k+1 \equiv 0 \pmod{9}, k+2 \equiv 0 \pmod{25}. Now, you apply the Chinese Remainder Theorem: k \equiv 548 \pmod{900}. So, 548,549,550 are divisible by 2^2,3^2,5^2 respectively. So, for the n-th case, let p_1,\ldots p_n be distinct primes, and solve the Chinese Remainder Theorem for 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.
    Thanks from virebbala90
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  3. #3
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    Re: number theory

    Thats wonderful
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