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Math Help - 2, 5, 8, 11, 14...

  1. #1
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    2, 5, 8, 11, 14...

    How can I prove that no square number can be in the sequence?

    2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...

    Is it to do with mod 5? And if so how would I do it
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Natasha
    How can I prove that no square number can be in the sequence?

    2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...

    Is it to do with mod 5? And if so how would I do it
    First you need find the general term of your sequence. Now it looks to me
    as though the sequence is:

    a_n=2+3n, n=0,1,2,..

    Which suggests you consider the squares modulo 3. Now any natural number
    may be written:

    N=3k+\rho,

    for some natural number k, and \rho= 0, 1\  \mbox{or}\  2. So

    N^2=9k^2+6k \rho + \rho^2.

    Therefore N^2(mod\ 3) is \rho^2(mod\  3), but

    <br />
\rho^2(mod\  3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2<br />
.

    So as a_n, n=0,1,2,.. is congruent to 2(mod\ 3) it cannot be a square for any
    natural number n

    RonL
    Last edited by CaptainBlack; February 12th 2006 at 12:35 PM.
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  3. #3
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    Quote Originally Posted by CaptainBlack
    First you need find the general term of your sequence. Now it looks to me
    as though the sequence is:

    a_n=2+3n, n=0,1,2,..

    Which suggests you consider the squares modulo 3. Now any natural number
    may be written:

    N=3k+\rho,

    for some natural number k, and \rho= 0, 1\  \mbox{or}\  2. So

    N^2=9k^2+6k \rho + \rho^2.

    Therefore N^2(mod\ 3) is \rho^2(mod\  3), but

    <br />
\rho^2(mod\  3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2<br />
.

    So as a_n, n=1,2,.. is congruent to 2(mod\ 3) it cannot be a square for any
    natural number n

    RonL
    Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Natasha
    Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
    Yes, but the +2 form is slightly more convienient when dealing with modular
    arithmetic.

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack
    Yes, but the +2 form is slightly more convienient when dealing with modular
    arithmetic.

    RonL
    I see. Thanks!
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