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Math Help - Proof by Induction?

  1. #16
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by Aquafina View Post
    lol ok, will this proof imply that n=3 is the only solution for n^4+n^3+n^2+n+1 being a perfect square?

    Yes!
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  2. #17
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    ok thanks

    now do you have any idea how i can show that

    (n^5 - 1)/(n-1) = k^2 (does not equal)

    for ONLY n=3

    where k is an integer
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  3. #18
    Super Member fardeen_gen's Avatar
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    Quote Originally Posted by Aquafina View Post
    ok thanks

    now do you have any idea how i can show that

    (n^5 - 1)/(n-1) = k^2 (does not equal)

    for ONLY n=3

    where k is an integer
    ?

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  4. #19
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    sorry, does equal k^2

    my bad
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  5. #20
    Super Member fardeen_gen's Avatar
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    The only way I see, is to prove that \frac{n^5 - 1}{n - 1}\neq k^2 for n\neq 3. We did the n > 3 part. I think you can make a separate thread for n < 2.
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  6. #21
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    Quote Originally Posted by fardeen_gen View Post
    The only way I see, is to prove that \frac{n^5 - 1}{n - 1}\neq k^2 for n\neq 3. We did the n > 3 part. I think you can make a separate thread for n < 2.
    I thought that n is a natural number... Because then by verifying for n = 1,2 you can see the expression's not a square. However n > 3 case, we already know the result. Thus n=3 is the only number that makes the expression a square...
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  7. #22
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    how come it only works with n=3? there must be some sort of reasoning behind it? because say if i had not known that n=3 is a solution (by inputting values into excel) then how would i have come about the answer?

    also, in the proof for n>3 this has been used:

    CASE I: n is even,
    (n^{2}+\frac{n}{2})^{2}<\frac{n^{5}-1}{n-1}< (n^{2}+\frac{n}{2}+1)^{2}

    CASE II: n is odd,
    (n^{2}+\frac{n-1}{2})^{2}<\frac{n^{5}-1}{n-1}< (n^{2}+\frac{n+1}{2})^{2}

    Where have the expressions for the boundaries come from? Are they related to n=3 in some way? Because if not, one could also say that this holds for n > 2 or n > 5 etc.
    Basically, whats the relationship between the boundaries used and n > 3?
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  8. #23
    Lord of certain Rings
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    Quote Originally Posted by Aquafina View Post
    Basically, whats the relationship between the boundaries used and n > 3?
    Did you expand fardeen's ingenious inequalities and see why they are true?
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  9. #24
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    sorry ill just paste the expression again in picture format

    CASE I: is even,
    <font size="3">

    i wasnt able to see the relationship sorry lol..

    i'm not an IIT student, i'm guessing u guys are

    could you tell me how they help?
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  10. #25
    Lord of certain Rings
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    Quote Originally Posted by Aquafina View Post
    sorry ill just paste the expression again in picture format

    CASE I: is even,
    <font size="3">

    i wasnt able to see the relationship sorry lol..

    i'm not an IIT student, i'm guessing u guys are

    could you tell me how they help?
    Aquafina,

    Take fardeen's 'n' even case, expand the equation (n^{2}+\frac{n}{2})^{2}- \frac{n^{5}-1}{n-1} and tell me what you get(is it positive or negative?). Then do the same thing with (n^{2}+\frac{n}{2}+1)^{2} - \frac{n^{5}-1}{n-1}  .

    Can you try it? Then tell us if you are stuck somewhere...
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  11. #26
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    ok i changed (n^5 - 1)/(n-1) to n^4 + n^3 + n^2 + n + 1

    then for the first part:

    -3/4 n^2 + n + 1

    and the second part to

    n^4 + n^3 + 9/4 n^2 + n + 1
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  12. #27
    Lord of certain Rings
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    Quote Originally Posted by Aquafina View Post
    ok i changed (n^5 - 1)/(n-1) to n^4 + n^3 + n^2 + n + 1

    then for the first part:

    -3/4 n^2 + n + 1
    I get -(\frac34  n^2  + n +  1)

    and the second part to

    n^4 + n^3 + 9/4 n^2 + n + 1
    I asked you to compute the difference:
    n^4 + n^3 + 9/4 n^2 + n  + 1 - (n^4+n^3+n^2+n+1) = \frac{5}{4}n^2

    Next step:
    Prove that -(\frac34  n^2  + n +  1) < 0 and \frac{5}{4}n^2 > 0

    \frac{5}{4}n^2 > 0 is obvious

    -(\frac34  n^2  + n +  1) < 0 is kinda obvious too..

    So now try the odd case
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  13. #28
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    okay..

    - (7/9 n^2 + 3/2 n + 3/4)

    and

    1/4 n^2 - 1/2 n - 3/4
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  14. #29
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    Quote Originally Posted by Aquafina View Post
    okay..

    - (7/9 n^2 + 3/2 n + 3/4)

    and

    1/4 n^2 - 1/2 n - 3/4
    Probably 9 was a typo, good work:

    I get (n^{2}+\frac{n-1}{2})^{2} - \frac{n^{5}-1}{n-1} = -\left(\frac74 n^2 + \frac32 n + \frac34\right) < 0

    Second one is right: (n^{2}+\frac{n+1}{2})^{2}- \frac{n^{5}-1}{n-1} = \frac{n^2 - 2n - 3}{4}. So when do you think n^2 - 2n - 3 > 0 holds?
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  15. #30
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    ah that's really good! (yes 9 was a typo sorry)

    how come only the upper boundary of n being odd is related to n > 3? i.e. with n^2 - 2n - 3 > 0

    Also, how would you come up with these equations from scratch!? I get how I worked back from fardeen's work to get the case...
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