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Math Help - two elementary number theory problem:

  1. #1
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    two elementary number theory problem:

    problem 1:
    show that for every positive integer  n , 7 divides  3^{2n+1}+2^{n+2}

    problem 2:
    show that if  n is any odd integer greater than 1 ,then  n^{5}-n is divisible by 80

    please help!
    Last edited by earthboy; October 2nd 2010 at 07:21 AM.
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  2. #2
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    Problem 2, as you have it written, is incorrect. For example, for n=2, n^5-n=2^5-2=32-2=30,
    and for n=4,
    n^5-n=4^5-4=1024-4=1020,
    neither of which is divisible by 80.
    Now, if you consider odd n greater that 1...

    --Kevin C.
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  3. #3
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    thank you!
    sorry! for writing the question incorrect.i have edited the question :-)
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  4. #4
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    Hello, earthboy!

    \text{[1] Show that for every positive integer }n\text{, 7 divides }3^{2n+1}+2^{n+2}

    We have: / 3^{2n+1} + 2^{n+2} \;\;=\;\;3\cdot3^{2n} + 2^2\cdot 2^n \;\;=\;\;3(3^2)^n + 4(2^n)

    . . . . . . =\;3\cdot9^n + 4\cdot2^n \;\;=\;\;3\cdot(7+2)^n + 4\cdot2^n

    . . . . . . =\;3\bigg(7^n + {n\choose1}7^{n-1}\!\cdot\!2 + {n\choose2}7^{n-2}\!\cdot\!2^2 + \hdots + {n\choose1}7\!\cdot\!2^{n-1} + 2^n\bigg) + 4\!\cdot\!2^n

    . . . . . . =\;3\bigg(7^n + {n\choose1}7^{n-1}\!\cdot\!2 + {n\choose2}7^{n-2}\!\cdot\!2^2 + \hdots + {n\choose1}7\!\cdot\!2^{n-1}\bigg) + 3\!\cdot\!2^n + 4\!\cdot\!2^n

    . . . . . . =\;3\bigg(7^n + {n\choose1}7^{n-1}\!\cdot\!2 + {n\choose2}7^{n-2}\!\cdot\!2^2 + \hdots + {n\choose1}7\!\cdot\!2^{n-1}\bigg) + 7\!\cdot\!2^n

    . . . . . . =\;3\!\cdot\!7\bigg(7^{n-1} + {n\choose1}7^{n-2}\!\cdot\!2 + {n\choose2}7^{n-3}\!\cdot\!2^2 + \hdots + {n\choose1}2^{n-1}\bigg) + 7\!\cdot\!2^n

    . . . . . . =\;7\!\cdot\!\bigg[3\bigg(7^{n-1} + {n\choose1}7^{n-2}\!\cdot\!2 + {n\choose2}7^{n-3}\!\cdot\!2^2 + \hdots + {n\choose1}2^{n-1}\bigg) + 2^n\bigg]

    There!
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  5. #5
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    Quote Originally Posted by Soroban View Post

    We have: / 3^{2n+1} + 2^{n+2} \;\;=\;\;3\cdot3^{2n} + 2^2\cdot 2^n \;\;=\;\;3(3^2)^n + 4(2^n)

    . . . . . . =\;3\cdot9^n + 4\cdot2^n \ldots
    It's a bit easier if you use congruence arithmetic starting from this point

    3\cdot9^n+4\cdot2^n\equiv3\cdot2^n+4\cdot2^n\equiv  7\cdot2^n\equiv0\pmod{7}
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  6. #6
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    Quote Originally Posted by earthboy View Post
    problem 2:
    show that if  n is any odd integer greater than 1 ,then  n^{5}-n is divisible by 80

    please help!
    Here's one thoughtless but valid approach

    Pari/GP:
    Code:
    for(i=1,79,if(i%2==1&&(i^5-i)%80!=0,print(i)))
    In fact the restriction "greater than 1" is unnecessary.
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  7. #7
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    When n = 2k -1 for some integer,  n^5 - n can be written as (2k-1)(4k^2 + 4k + 2)(2k)(2k-2). The last three factors are even and 4 divides either 2k or 2k-2. So 16 divides the product. If 5 does not divide any among 2k-1 and 2k-2 and 2k, then it must leave remainder 2 or 3 when it divides n. The second factor is (n^2 + 1). If n=5m+2 or 5m+3, then (n^2 + 1) is 25m^2 + 20m + 5 and 25m^2 + 30m + 10 respectively.
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  8. #8
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    problem 2:
    show that if n is any odd integer greater than 1 ,then n^5-n is divisible by 80
    Actually in a more closer look n^5-n is divisible by 240 (for any odd n).
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  9. #9
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    Or a proof by induction on n: when n= 0, 3^{2n+1}+ 2^{n+2}= 3^1+ 2^2= 3+ 4= 7 which is divisible by 7.

    Assume that, for some k, 3^{2k+1}+ 2^{k+2} is divisible by 7. That is, 3^{2k+1}+ 2^{k+2}= 7j for some integer j.

    Then 3^{2(k+1)+ 1}+ 2^{(k+1)+ 2}= 3^{2k+ 1+ 2}+ 2^{k+2+ 1}= 9(3^{2k+1})+ 2(2^{k+1}) = 2(3^{2k+1}+ 2^{2k+1}+ 7(3^{2k+1})= 2(7j)+ 7(3^{2k+1})= 7(2j+ 3^{2k+1}) is divisible by 7.

    By induction on n, then, 3^{2k+1}+ 2^{k+ 2} is divisible by 7 for all n.
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