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    (Original post by DeanK22)
    TSR was the cause for quite literally doing nothing in terms of school work and I thought this would have affected my exams (considering I had done no homework in any of my subjects) and I need three a for Oxford. Considering how my exams have recently gone though I found that you can quite literally learn whole modules in one or three days - subject specific - ergo TSR time.
    Lol fair enough.
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    (Original post by DeanK22)
    This is the same person :rolleyes: ...
    lmao!
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    Prettier rep though.
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    (Original post by The Muon)
    lmao!
    Anymore wall challenges?
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    how about something simple, criticise the logical paradox in the statement:
    "This statement is false"

    lol dean you made a new accout because your rep was low :rofl:
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    Quickie.

    Sir Gustaff is happy when smoking ciggareetes and drinking wine. So happy his happyness is proprtional to the square of the number of bottels of wine he has in one day multiplied by the packets of ciggarettes he has in one day. A bottle of wine costs 3 pounds, ciggarettes cost 2 pound. He has an allowance of 100 pounds. What amount of alcohol and ciggarettes should Gustaff acquire to achive greatest happyness?
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    question
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    (Original post by DeanK22)
    question
    answer :p:
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    (Original post by The Muon)
    answer :p:
    Prove that  \sqrt{2} + \sqrt{3} is algebraic
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    not a good problem
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    (Original post by DeanK22)
    Prove that  \sqrt{2} + \sqrt{3} is algebraic
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    not a good problem
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    it is a solution of  x^4 - 10x^2 + 1 = 0 . i think this suffices...
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    (Original post by GHOSH-5)
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    it is a solution of  x^4 - 10x^2 + 1 = 0 . i think this suffices...
    A quaint paper 2 question;

    4) Prove there are infinitely many integers [distinct] that are positive such that  x^2 + y^3 is divisible by x^3 + y^2
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    (Original post by DeanK22)
    A quaint paper 2 question;

    4) Prove there are infinitely many integers [distinct] that are positive such that  x^2 + y^3[/late] is divisible by [latex]x^3 + y^2
    is this from a BMO-2?
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    (Original post by GHOSH-5)
    is this from a BMO-2?
    correct
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    (Original post by DeanK22)
    correct
    out of curiosity, do you actually have a solution to the problem?
  15. Offline

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    (Original post by GHOSH-5)
    out of curiosity, do you actually have a solution to the problem?
    I do
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    (Original post by JohnnySPal)
    I know I'm being hick here, but why doesn't the solution set {(x,x) : x \in \mathbb{N}}
    Oh. Yeah I made that mistake at the first time I read it.

    (x,y) are such that  x \ne y - distinct.
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    (Original post by DeanK22)
    I do
    a hint by any chance?
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    (Original post by GHOSH-5)
    a hint by any chance?
    It would considereably give the game away as the hint is the solution.

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    Don't open three spoilers below before thinking hard
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    Why wlog can you assume something about (x,y) ?
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    assume y > x
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    y = kx
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    Solution;

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    Let y = kx  \Rightarrow \frac{x^2 + y^3}{x^3 + y^2} = \frac{x^2 + k^3x^3}{x^3 + k^2x^2} =  \frac{k^3x + 1}{x + k^2} = \frac{k^3x + k^5 - k^5 + 1}{x + k^2} = k^3 - \frac{k^5 - 1}{x + k^2} It is clear that  k^5 - 1 = n(x + k^2) for some n. You can take it that  \frac{k^5 - 1}{x+k^2} = 1 so x must be  k^5 - k^2 - 1 and therefore  y = k^6 - k^3 - k
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    (Original post by DeanK22)
    Solution;

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    Let y = kx  \Rightarrow \frac{x^2 + y^3}{x^3 + y^2} = \frac{x^2 + k^3x^3}{x^3 + k^2x^2} =  \frac{k^3x + 1}{x + k^2} = \frac{k^3x + k^5 - k^5 + 1}{x + k^2} = k^3 - \frac{k^5 - 1}{x + k^2} It is clear that  k^5 - 1 = n(x + k^2) for some n. You can take it that  \frac{k^5 - 1}{x+k^2} = 1 so x must be  k^5 - k^2 - 1 and therefore  y = k^6 - k^3 - k
    nice. i was at a point of  k^3 -\frac{k^5-1}{x+k^2}
    i didnt really think to let the fraction equal one

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