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Math Help - Geometric and quadratic means

  1. #1
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    Geometric and quadratic means

    I suspect that there is no pair of distinct integers a, b for which the geometric mean and the quadratic mean (root mean square) are both integers, but I am getting nowhere trying to prove it. Any ideas?
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    Bar0n janvdl's Avatar
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    I am sorry if i just dont understand, but could you please be a little more clear?
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    Quote Originally Posted by icebag View Post
    I suspect that there is no pair of distinct integers a, b for which the geometric mean and the quadratic mean (root mean square) are both integers, but I am getting nowhere trying to prove it. Any ideas?
    By Quadradic Mean I think you mean, Generalized Power Mean which exponent 2.

    Then if you choose a=b integers.
    Then,
    QM(a,b)=a=b --> integer.
    GM(a,b)=a=b --> integer.

    What is the problem?
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    *distinct* integers, so a not equal to b
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    Quote Originally Posted by janvdl View Post
    I am sorry if i just dont understand, but could you please be a little more clear?
    Let a and b be unequal integers.

    The geometric mean is g=sqrt(a*b).

    The root mean square (or quadratic mean) is q=sqrt(0.5*(a^2+b^2)).

    I want to prove that g and q cannot both be integers.

    (Or, of course, find values of a and b such that g and q are both integers.)
    Last edited by icebag; May 4th 2007 at 02:00 PM. Reason: My silly error!
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    Bar0n janvdl's Avatar
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    This is as far as i get...

    g = sqr(ab)
    q = sqr(0,5a^2 + 0,5b^2)

    g = q

    sqr(ab) = sqr(0,5a^2 + 0,5b^2)

    ab = 0,5a^2 + 0,5b^2

    2ab = a^2 + b^2

    0 = a^2 - 2ab + b^2

    0 = (a - b)^2
    Last edited by janvdl; May 4th 2007 at 01:00 PM. Reason: Further Simplification...
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    Quote Originally Posted by janvdl View Post
    This is as far as i get...

    g = sqr(ab)
    q = sqr(0,5a^2 + 0,5b^2)

    g = q

    sqr(ab) = sqr(0,5a^2 + 0,5b^2)

    ab = 0,5a^2 + 0,5b^2

    2ab = a^2 + b^2

    0 = a^2 - 2ab + b^2

    0 = (a - b)^2
    You cannot do that.
    Because you are assuming that GM and QM are equal.

    They are not necessarily equal.
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  8. #8
    Bar0n janvdl's Avatar
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    Im sorry PerfectHacker, im not on varsity yet, im only trying to help.

    But what if we do ASSUME them to be equal? Cant it aid us in finding a and b's value?
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    Bar0n janvdl's Avatar
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    Well then im working in reverse but who cares...

    Quote Originally Posted by janvdl View Post
    g = sqr(ab)
    q = sqr(0,5a^2 + 0,5b^2)

    g = q

    sqr(ab) = sqr(0,5a^2 + 0,5b^2)

    ab = 0,5a^2 + 0,5b^2

    2ab = a^2 + b^2

    0 = a^2 - 2ab + b^2

    0 = (a - b)^2
    So if we take the sqr root on both sides:
    0 = a - b
    a = b

    So for g to equal q:
    a must equal b.

    So if a and b are distinct, my logic tells me that g will not equal q.

    I dont know if everything is right, but maybe it can give you a clue.
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  10. #10
    Bar0n janvdl's Avatar
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    Hey icebag you edited your post...

    The first one didnt say to prove that g and q were integers, but to prove them equal...

    Ok, so my solution is way off, sorry.
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  11. #11
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    Quote Originally Posted by icebag View Post
    Let a and b be unequal integers.

    The geometric mean is g=sqrt(a*b).

    The root mean square (or quadratic mean) is q=sqrt(0.5*(a^2+b^2)).

    I want to prove that g and q cannot both be integers.

    (Or, of course, find values of a and b such that g and q are both integers.)
    I can reduce these conditions to a single equation. Perhaps ThePerfectHacker can take it from where I have to leave it.

    Let b > a. Then b = a + n where n is some positive integer. Then
    q = sqrt{(1/2)*(a^2 + a^2 + 2an + n^2)}

    q = sqrt{a^2 + an + (1/2)n^2}

    For this to be an integer n must at least be an even number. Thus a and b are either both even, or both odd.

    Case 1: a, b are both odd.
    Let a = 2x + 1
    Let b = 2y + 1
    (x, y both integers)

    Then
    q = sqrt{(1/2)(4x^2 + 4x + 1 + 4y^2 + 4y + 1)}

    q = sqrt{2(2x^2 + 2x + 2y^2 + 2y + 1)}

    But for q to be an integer 2x^2 + 2x + 2y^2 + 2y + 1 = 2(x^2 + x + y^2 + y) + 1 must be an even number, which is false.

    Thus a, b are not both odd.

    Case 2: a = 0 (mod 4) and b = 2 (mod 4)
    Let a = 4x
    Let b = 4y + 2
    (x, y both integers)

    Then
    q = sqrt{(1/2)[(4x)^2 + (4y + 2)^2]}

    q = sqrt{2(4x^2 + 4y^2 + 4y + 1)}

    But for q to be an integer 4x^2 + 4y^2 + 4y + 1 = 2(2x^2 + 2y^2 + 2y) + 1 must be an even number, which is false.

    Thus etc.

    Case 3: a = 0 (mod 4) and b = 0 (mod 4)
    Let a = 4x
    Let b = 4y

    Then g = 4*sqrt{xy}
    Then q = 4*sqrt{(1/2)(x^2 + y^2)}

    If g and q are to be integers then this implies that the geometric mean and quadratic means of x and y are integers. This is to say that we may divide a and b by 4 and still have a pair of numbers that works. So this is not a case in and of itself, but can always be reduced to one of the other cases. Since no other case works, a and b must then be of the form...

    Case 4: a = 2 (mod 4) and b = 2 (mod 4)
    Let a = 4x + 2
    Let b = 4y + 2
    (x, y are integers)

    Then
    g = sqrt{(4x + 2)(4y + 2)} = 2*sqrt{(2x + 1)(2y + 1)}
    The condition that g is an integer implies that 2x + 1 is a factor of 2y + 1 of the form:
    2y + 1 = (2x + 1)*z^2
    where z is some integer.

    Thus
    q = sqrt{(1/2)[(2x + 1)^2 + (1/2)[(2x + 1)z^2]^2}

    q = (2x + 1)*sqrt{(1/2)(z^4 + 1)}

    If q is to be an integer then so is c = sqrt{(1/2)(z^4 + 1)}.

    Thus we need to solve the equation:
    2c^2 - z^4 = 1 where c, z are integers.

    If a solution to this equation exists we may pick a value of x and then generate a value for y, and thus a and b.

    I can't solve this (or prove it has no solution). Perhaps TPH can do it since he seems to have studied these types of equations.

    -Dan
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  12. #12
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    I can't follow the reasoning used in this bit:


    Quote Originally Posted by topsquark View Post
    Then
    g = sqrt{(4x + 2)(4y + 2)} = 2*sqrt{(2x + 1)(2y + 1)}
    The condition that g is an integer implies that 2x + 1 is a factor of 2y + 1 of the form:
    2y + 1 = (2x + 1)*z^2
    where z is some integer.

    The product (2x + 1)(2y + 1) can be a perfect square without either term being a factor of the other: e.g. 9 x 25 has square root 15. Am I missing something?
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  13. #13
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    Here.

    (See where you can spot where I use the fact that they are distinct in the proof).
    Attached Thumbnails Attached Thumbnails Geometric and quadratic means-picture24.gif  
    Last edited by ThePerfectHacker; May 5th 2007 at 06:16 PM.
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  14. #14
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by icebag View Post
    I can't follow the reasoning used in this bit:





    The product (2x + 1)(2y + 1) can be a perfect square without either term being a factor of the other: e.g. 9 x 25 has square root 15. Am I missing something?
    Ah. I missed that case. Oh well, it WAS 3:30 in the morning after all...

    -Dan
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  15. #15
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    Thank you TPH. That is very neat.

    (Interestingly, I had got to the x^4 - y^4 = z^2 form myself, but I just didn't know that it had no integer solutions.)
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