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    prove that

    prove that the product of five successive natural numbers can't be a perfect square.
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    Quote Originally Posted by archimedes81 View Post
    prove that the product of five successive natural numbers can't be a perfect square.
    Let the five numbers be: n,\ n+1,\ n+2,\ n+3,\ n+4 their product is divisible by n, and if n>4 is not divisible by n^2 as that would require that 1\ 2,\ 3 or 4 were divisible by n.

    The cases n=1,2,3,4 can be eliminated by direct calculation.

    CB
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    Moo
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    Hi,
    Quote Originally Posted by CaptainBlack View Post
    Let the five numbers be: n,\ n+1,\ n+2,\ n+3,\ n+4 their product is divisible by n, and if n>4 is not divisible by n^2 as that would require that 1\ 2,\ 3 or 4 were divisible by n.

    The cases n=1,2,3,4 can be eliminated by direct calculation.

    CB
    I had a thought on this problem... And started like you. But let's imagine n=pq.
    Then, we can have p divides 2 (p=2) and q divides 4 (q=4). But yet n=8 and doesn't divide 1,2,3,4.

    Whereas if we consider the consecutive numbers n-2,n-1,n,n+1,n+2, we have their product equal to :
    n(n^2-1)(n^2-4)=n(n^4-5n^2+4)
    And from this, it should be clear that n has to divide 4.


    BUT there is still a problem. What happens if n is a square ? Or if n contains a square in its factors ???
    We have to reconsider the equation
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    Quote Originally Posted by Moo View Post
    Hi,

    I had a thought on this problem... And started like you. But let's imagine n=pq.
    Then, we can have p divides 2 (p=2) and q divides 4 (q=4). But yet n=8 and doesn't divide 1,2,3,4.
    Consider the prime decomposition of n=p_1^{k_1} ... p_r^{k_r}

    We have (n+1)(n+2)(n+3)(n+4) is divisible by n (if the product of the five consecutive numbers is a perfect square). So p_i divides one of 2,3 or 4, and the only possibilities this leaves for n are 2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24 and these can be eliminated by direct computation (I could add additional justification here but the reader should be able provide that for themselves - if I have got the argument right).

    That should work, or have I still missed something?

    CB
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    Moo
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    Quote Originally Posted by CaptainBlack View Post
    Consider the prime decomposition of n=p_1^{k_1} ... p_r^{k_r}

    We have (n+1)(n+2)(n+3)(n+4) is divisible by n (if the product of the five consecutive numbers is a perfect square). So p_i divides one of 2,3 or 4, and the only possibilities this leaves for n are 2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24 and these can be eliminated by direct computation (I could add additional justification here but the reader should be able provide that for themselves - if I have got the argument right).

    That should work, or have I still missed something?

    CB
    This works, yes ^^

    But now, you haven't considered the situation where n has a square in its factor (or even if it's a square itself).
    Because then we wouldn't need n to divide (n+1)(n+2)(n+3)(n+4)
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