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Math Help - (n!(j)+1,n!(j)+1)=1 for 1<=i<j<=n

  1. #1
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    (n!(j)+1,n!(j)+1)=1 for 1<=i<j<=n

    Please i need this bt tomorrow morning, and I have been staring at it for 5 hours.
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  2. #2
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    Quote Originally Posted by Jer2145 View Post
    Please i need this bt tomorrow morning, and I have been staring at it for 5 hours.
    care to explain a bit? what is n!(j)?
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  3. #3
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    Quote Originally Posted by Jer2145 View Post
    Please i need this bt tomorrow morning, and I have been staring at it for 5 hours.
    1. Please state the problem in the text of the thread, not just in the title.

    2. Please state the problem correctly.

    3. I'm guessing that (n!(j)+1,n!(j)+1)=1 for 1<=i<j<=n means gcd(in! + 1, jn! + 1) = 1.

    4. If d divides in! + 1 and jn! + 1, then d also divides their difference, which is (ji)n!. But all the prime factors of the product (ji)n! must lie between 1 and n. It follows that any prime factor of d must lie between 1 and n. So d cannot divide in! + 1. Therefore d has no prime factors, so d must be 1.
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