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    Hello,

    I was just attempting a question, which you can find here. I think I've got a solution, but I was hoping that someone would be kind enough to check it :o:
    (Original post by Question 3, BMO-2, 1998)
    Suppose x, y, z are positive integers satisfying the equation

     \dfrac{1}{x} - \dfrac{1}{y} = \dfrac{1}{z},

    and let h be the highest common factor of x, y, z.
    Prove that hxyz is a perfect square.
    Prove also that h(y-x) is a perfect square.
    Attempt
    Let  x = hX, y = hY, z = hZ, and so there is no positive integer greater than one that divides all three  X, Y, Z (Property A). Therefore

     \dfrac{1}{X} - \dfrac{1}{Y} = \dfrac{1}{Z}

     \iff ZY - ZX = XY \ \ \ (\ast )

    Therefore we find that  Z|XY, Y|XZ, X|YZ.

    Suppose that  \mathrm{gcd}(X,Y) = a, \mathrm{gcd}(X,Z) = b, \mathrm{gcd}(Y,Z) = c.

    Let:
     X = ax_1, Y = ay_1 , such that  \mathrm{gcd}(x_1, y_1) = 1
     X = bx_2, Z = bz_1 , such that  \mathrm{gcd}(x_2, z_1) = 1
     Y = cy_2, Z = cz_2 , such that  \mathrm{gcd}(y_2, z_2) = 1

    Let us show that \mathrm{gcd}(x_1, x_2) = \mathrm{gcd}(y_1, y_2) = \mathrm{gcd}(z_1, z_2) = 1.

    Suppose \mathrm{gcd}(x_1, x_2) = f. Note that  f|X.  X|YZ \iff ax_1|ay_1Z \implies x_1|Z \implies f|Z. Similarly  X|YZ \iff bx_2|bz_1Y \implies x_2|Y \implies f|Y.*

    But due to Property A, f must be one, and so \mathrm{gcd}(x_1, x_2) = 1. A similar argument goes for \mathrm{gcd}(y_1, y_2) and \mathrm{gcd}(z_1, z_2).

    We can also show quite easily that  a, b, c are also pairwise coprime: suppose two of them have a common factor greater than one; then this divides all three of X, Y, Z, which is a contradiction.

    Therefore,  X = ax_1 = bx_2 , and so  x_1 = b, x_2 = a \implies X = ab. We can similarly show that  Y = ac, Z = bc.*

    Therefore  hxyz = h^4XYZ = (h^2abc)^2 . Q.E.D. 1.

    Going back to equation  (\ast ) , we find that

     ZY - ZX = XY \iff Z(Y-X) = XY \iff Y-X = a^2

    And therefore  h(y-x) = h^2(Y-X) = (ha)^2 . Q.E.D. 2.
    I think that works, but I was most unsure at the two paragraphs I've ended with a small asterisk*.

    Thanks in advance for any help :smile:
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    Wiki Support Team
    It all looks alright to me.
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    Thanks very much.
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    Hmm. I was trying question 4 of this paper, and I think I've sorted out the first part, and although I'm pretty sure the route I'm trying for the second part isn't the intended one (involving a little calculus), I'm not sure why it's not working. If anyone help out, I'd appreciate it.
    (Original post by Question 4, BMO-2, 1998)
    Find a solution of the simultaneous equations

     xy + yz + xz = 12

     xyz = 2 + x + y+ z

    in which all of x, y, z are positive and prove that it is the only such solution.
    Show that a solution exists in which x, y, z are real and distinct.
    First Part Attempt
    For positive numbers x, y, z, by AM-GM

     4 = \dfrac{xy+yz+xz}{3} \geq \sqrt[3]{x^2y^2z^2}

     \implies xyz \leq 8 \iff x+y+z \leq 6

     \therefore (x+y+z)^2 \leq 36

     \iff x^2+y^2+z^2 \leq 12 = xy+yz+xz

     \therefore 2x^2+2y^2+2z^2-2xy-2yz-2xz \leq 0

     \iff (x-y)^2 + (y-z)^2 + (x-z)^2 \leq 0

    But since these terms can't be negative, they must sum to zero, meaning they are all equal, and hence we find that x = y = z = 2 is the only possibility for positive x, y, z
    Attempt at Second Part
    If we consider the polynomial

     P(w) = w^3 - cw^2+12w-(c+2) , where  c = x+y+z, then x, y, z are the roots of  P(w)=0. For the roots to be distinct we need the turning points (two distinct ones) of this cubic to essentially be either side of the 'w-axis'.

    So the turning points are at  P'(w)=0 and we find that these are at  w = \dfrac{c \pm \sqrt{c^2-36}}{3} ( c^2 > 36), and so we require the following two inequalities:

     P\left(\dfrac{c -\sqrt{c^2-36}}{3}\right) > 0 and  P\left(\dfrac{c +\sqrt{c^2-36}}{3}\right) < 0

    And after a lot of messy algebra (perhaps some of it is wrong), we end up with:

     -2c^3+2c^2\sqrt{c^2-36} + 81c-72\sqrt{c^2-36}-54 > 0

    and

     -2c^3-2c^2\sqrt{c^2-36} + 81c+72\sqrt{c^2-36}-54 < 0

    Then I attempt to combine the two inequalities and get

    4c^2\sqrt{c^2-36}-144\sqrt{c^2-36} &gt; 0 \iff (c^2-36)\sqrt{c^2-36} &gt; 0 which seems to imply that we have distinct real roots to P(w) for any c > 6, c < -6. But this clearly isn't true, for example checking c = 7 and plotting the graph, there is only one solution (near 5). :confused:
    Can anyone lend a hand? Or perhaps suggest an alternate method? I can't seem to spot a nice algebraic method for this part. Thanks again for any help! :o:
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    Sorry, I misread earlier, deleted as soon as I realised. For the last part, guess a value for one of them (sensible guess!) and see whether you can find a solution for the others
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    (Original post by SimonM)
    Sorry, I misread earlier, deleted as soon as I realised.
    No problem! :smile:
    For the last part, guess a value for one of them (sensible guess!) and see whether you can find a solution for the others
    Cheers.
    Spoiler:
    Show
    The solution  \left( -\dfrac{3}{2}, -5, -\dfrac{9}{13}\right) works fine. :smile:
    I'm wondering what you meant exactly by 'sensible'.
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    I'm also curious. To me the obvious "nice" guess was x=-1, but that ends up with y = -1, z = 11/2.
    x = -2 gives 3 distinct roots, even though the other 2 roots aren't particularly nice.

    Still, trial and error is a perfectly reasonable approach here.
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    (Original post by DFranklin)
    To me the obvious "nice" guess was x=-1, but that ends up with y = -1, z = 11/2.
    Indeed. :erm:
    I'm also curious.
    Hmm. Considering the route I originally went for, I 'cheated' and used a computer for a little bit: I decided to plot a graph of the cubic discriminant against the value of c (some horrible quartic I think), and it showed that we get three distinct roots if x+y+z is somewhere in between -7.5 and -8.5, or something like that....
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    One quick question on BMO-2 2009, whether the following method works :confused:
    (Original post by Question 1, BMO-2, 2009)
    Find all solutions in non-negative integers a, b to  \sqrt{a}+\sqrt{b} = \sqrt{2009}
    Spoiler:
    Show
    \sqrt{a}+\sqrt{b} = \sqrt{2009}\implies 2\sqrt{ab} = 2009-a-b

     \implies 4ab = a^2+b^2+2009^2-4018a-4018b+2ab

     \iff a^2 + a(-4018-2b)+(b-2009)^2 = 0

     \therefore a = \dfrac{2b+4018\pm \sqrt{4(b+2009)^2-4(b-2009)^2}}{2} = b+2009 \pm 14\sqrt{41b}

    But the plus-or-minus sign can't be a plus, otherwise a &gt; 2009, which is not possible (as we are taking positive square roots in the original equation), and so therefore

     a = b+2009 - 14\sqrt{41b}

    Therefore b is a square multiple of 41 (as otherwise a is not an integer), and so is a, as the equation is symmetric, but we note that  a, b\leq 2009 =41 \times 7^2

    Therefore we find that  a, b = 0(41), 1(41), 4(41), 9(41), \dots 49(41) essentially.

    So we end up with ordered pairs of solutions  \boxed{(0, 2009), (41, 1476), (164, 1025), (369, 656)}
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    I have checked too closely but the method looks fine
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    (Original post by SimonM)
    I have checked too closely but the method looks fine
    :woo:
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    When I (first) did this your method of squaring repeatadly was my option but glancing at it now this is far quicker;

     \sqrt{a}+\sqrt{b} = \sqrt{2009} = 7\sqrt{41} and by leting  a = 41c^2 , b=41d^2 we are left with the far simpler problem c + d = 7
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    (Original post by DeanK22)
    When I (first) did this your method of squaring repeatadly was my option but glancing at it now this is far quicker;

     \sqrt{a}+\sqrt{b} = \sqrt{2009} = 7\sqrt{41} and by leting  a = 41c^2 , b=41d^2 we are left with the far simpler problem c + d = 7
    I noticed that as well, but I don't think that strictly proves that a and b are of the form  41k^2; it doesn't really show why they can't be a number not of that form :erm: at least, I don't think it does.
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    by squaring once you can see that either a and b have at least 1 common prime factor with an odd exponent (and possibly other prime factors that are not common with even exponents) or a and b are perfect squares with no common factors. Clearly it must be the former say a = c^2k and b=d^2k. The result follows
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    sorry for the mini hijack but can anyone give me hints on this.

    find the minimum value of |12^m-5^n|
    all i've spotted is
    Spoiler:
    Show
    that 5=2+3 12=2*2*3 so i did the binomial expansion of (2+3)^n but i couldn't see where to go. also n>=m


    hints?
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    (Original post by Totally Tom)
    sorry for the mini hijack but can anyone give me hints on this.
    No worries
    find the minimum value of |12^m-5^n|
    all i've spotted is
    Spoiler:
    Show
    that 5=2+3 12=2*2*3 so i did the binomial expansion of (2+3)^n but i couldn't see where to go. also n>=m


    hints?
    Hmm. Well 12-5 = 7. And you should be able to show that  |12^m - 5^n| \not= 2, 3, 4, 5, 6 for all m and n (I'm guessing we're talking positive integers here), so the question is really are there any pairs (m, n) such that  |12^m - 5^n| = 1 ?

    Further hint:
    Spoiler:
    Show
    Consider the two cases  12^m - 5^n = 1 and  5^n - 12^m = 1 . In the first, can you show that m must be even (actually a multiple of 4)? Similarly, in the second, can you show that n must be even? Then after a little rearrangement, think difference of two squares.
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    (Original post by DeanK22)
    by squaring once you can see that either a and b have at least 1 common prime factor with an odd exponent (and possibly other prime factors that are not common with even exponents) or a and b are perfect squares with no common factors. Clearly it must be the former say a = c^2k and b=d^2k. The result follows
    Ah yes. This is true. :smile:
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    Hmm I was trying BMO-2, 1994, Q1, and I've met a block - I'm probably missing something obvious :confused:
    (Original post by Question 1, BMO-2, 1994)
    Find the first integer n > 1 such that the average of

     1^2, 2^2, 3^2, \dots n^2

    is itself a perfect square.
    Spoiler:
    Show
    Well the average of the first n squares is:

     \dfrac{(2n+1)(n+1)}{6}

    So we're essentially looking for solutions to the equation:

     (2n+1)(n+1) = 6m^2

    Note that n+1 and 2n+1 are coprime, and also that 2n+1 is odd, so we have either the following cases:

     2n+1 = a^2, n+1 = 6b^2 \ \ \ \ \ \ \ \ \ 2n+1 = 3a^2, n+1=2b^2

    Let's dismiss the first case.

     n\equiv -1 \pmod{6} \iff 2n+1 \equiv -1 \pmod{6} , but -1 is not a quadratic residue modulo 6, so this is not possible.

    So  n+1 = 2b^2, 2n+1 = 3a^2 \iff 3a^2 = 4b^2 - 1 = (2b+1)(2b-1)

    So either  2b+1 = 3x^2, 2b-1 = y^2 or  2b+1 = x^2, 2b-1 = 3y^2 , x and y are both odd.

    Now let us dismiss the latter case:  x^2 - 3y^2 = 2 \implies x^2 \equiv 2 \pmod{3} , but 2 is not a quadratic residue modulo 3, so this is not possible.

    So essentially we are left to find the integer solutions of  3x^2 - y^2 = 2 , and I can't seem to do this (other than x = 1, y = 1, which results in n = 1).

    I suppose I could look down the route of a massive case bash involving 3(x+1)(x-1) = (y+1)(y-1), but is there a nice approach?
    Any hints would be great :smile:
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    Just grinding through options is doable (and what I think most people who get it out under time pressure end up doing - it's what I did).

    There's some discussion of this question here: http://www.thestudentroom.co.uk/show...=475982&page=2
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    that is a quadratic diophantine equation that you can solve. Have you come across quadratic diophantine equations before?

    infact, here is a site with step-by-step method with the above problem;

    http://www.alpertron.com.ar/QUAD.HTM
 
 
 
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