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Math Help - Revision related number theory question(s)

  1. #1
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    Revision related number theory question(s)

    I don't have a clue to even start the following question so any pointers would be welcome!

    Let n be an integer with n greater than or equal to 2. Show that n! + 2 is even.
    Now show that m divides n! + m, for all m such that 2<m<n (here it is a less than or equal to sign) and show why n! +m is composite.
    Use this to find a sequence of n-1 consecutive integers (that is a sequence a2, a3,...,an, with ai+1=ai +1 and all ai composite).
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by schteve View Post
    I don't have a clue to even start the following question so any pointers would be welcome!

    Let n be an integer with n greater than or equal to 2. Show that n! + 2 is even.
    Hint: n! = n(n - 1)(n - 2) \cdots 3 \cdot {\color{red}2} \cdot 1

    Now show that m divides n! + m, for all m such that 2<m<n (here it is a less than or equal to sign) and
    Again, m will be one of the factors of n! (i hope by now you see why. if not, think about it some more)

    show why n! +m is composite.
    by the above, it would be divisible by m > 2. hence, it cannot be prime, it must be composite.


    Use this to find a sequence of n-1 consecutive integers (that is a sequence a2, a3,...,an, with ai+1=ai +1 and all ai composite).
    take the sequence \{ a_i \}_{i = 2}^{n} =  n! + i

    do you see why this works?
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  3. #3
    Super Member Bacterius's Avatar
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    n! with n greater than 1 is even, because it has 2 as a factor. I think you should be ok for the first question.

    Since 2 < m < n, m divides n! (because m is a factor of n! as m < n).
    Thus you can say that n! + m = m \times \frac{n!}{m} + m = m(\frac{n!}{m} + 1) with \frac{n!}{m} \in \mathbb{N}. Thus, n! + m is composite because it has m as a factor, and we know that m > 2.
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  4. #4
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    Thanks for the help. Yeah its all clicked into place after just understanding the 1st stage.
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