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Solve this Maths question, or else, you aren't good enough for Oxbridge Watch

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    (Original post by illusionz)
    But that is only 1 equation...
    No, it's 2. The only thing is that the first implies the second. It was an example as to how yo can get more solutions. In the case of the OP, the equations are quadratic and cubic so multiple solutions likely exist that are not just rearangments of 1.
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    (Original post by JAIYEKO)
    We are not in primary school-.
    Well you need to go back there if you think your equations have an integer solution.
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    Has anyone ever though that Legendre could be wrong?

    After all, JAIYEKO is "UK U17s Best Mathematician".

    JAIYEKO, contact the papers and let them know the proof on
    http://www.rsabey.pwp.blueyonder.co.uk/maths/sumof3squares.html is incorrect because you have found a case where it doesn't work!!
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    (Original post by rayquaza17)
    Has anyone ever though that Legendre could be wrong?

    After all, JAIYEKO is "UK U17s Best Mathematician".

    JAIYEKO, contact the papers and let them know the proof on
    http://www.rsabey.pwp.blueyonder.co.uk/maths/sumof3squares.html is incorrect because you have found a case where it doesn't work!!
    You don't need any fancy theorems to prove the lack of integer solutions.

    If x^2+y^2+z^2=39, then x,y,z can be 0,1,4,9,16,25,36

    As 3*9=27<39, at least one of them must be 16 or 25 or 36. If x^2 is 36 then y^2+z^2=3 which is impossible.

    If x^2=25 then y^2+z^2=14 which is again impossible.

    If x^2=16 then y^2+z^2=23, and 2*9=18<23 so at least one of y^2 or z^2 is 16, so if y^2=16 then z^2=7 which is impossible.
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    (Original post by Hunarench95)


    For anyone that's interested ^
    I tried to cheat around the math by trying to find a 3D graph generator.
    Do you know any 3D graph generators that would work for PC I have windows 7. It would be very helpful to me for other things too.
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    (Original post by MathMeister)
    I tried to cheat around the math by trying to find a 3D graph generator.
    Do you know any 3D graph generators that would work for PC I have windows 7. It would be very helpful to me for other things too.
    I'm not sure of the price, but 'Maple' is an all in one mathematical program that plots 3d graphs - and even makes a fair attempt at 4d!! (If you're a university student, it may be subsidised)


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    It's a shame the math forum has come to this. I remember when it was a thriving metropolis during exam time. Alas, that was an age ago having viewed this thread.
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    You hear that people, if you can't do it you're not good enough for oxbridge...sorry English students :/
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    (Original post by Munrot07)
    You hear that people, if you can't do it you're not good enough for oxbridge...sorry English students :/
    A better answer would be if you can do it you are not good enough for oxbridge, because if you can do it you made a mistake.
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    (Original post by Spairos)
    But I am talking about the given equations of this question. The second equation of yours is the same as the first one.
    Consider solutions in integers to

    x^2+y^2=25
    x-y=1

    2 equations, both very different, but two distinct solutions.


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    (Original post by james22)
    No, it's 2. The only thing is that the first implies the second. It was an example as to how yo can get more solutions. In the case of the OP, the equations are quadratic and cubic so multiple solutions likely exist that are not just rearangments of 1.
    The second implies the first as well unless we are working in a modular system (not the case here). Your example is not hugely instructive.


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    (Original post by LightBlueSoldier)
    The second implies the first as well unless we are working in a modular system (not the case here). Your example is not hugely instructive.


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    It was an example to disprove the idea that n equations in n unknowns means 1 solution.
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    (Original post by james22)
    It was an example to disprove the idea that n equations in n unknowns means 1 solution.
    But your example does not involve n equations. It involves one equation.


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    (Original post by LightBlueSoldier)
    But your example does not involve n equations. It involves one equation.


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    If you have a system of n equations where the nth equation is implied by the other n-1 (basically any system with infinite solutions), do you really consider it anything other than a system of n equations?
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    (Original post by james22)
    If you have a system of n equations where the nth equation is implied by the other n-1 (basically any system with infinite solutions), do you really consider it anything other than a system of n equations?
    Typically when doing linear algebra if you know a piece of information is redundant you discard. Typically you search for minimal systems as this tells you lots of useful things.

    Tldr yes I would regard it as a system of (n-1) equations.


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    (Original post by LightBlueSoldier)
    Typically when doing linear algebra if you know a piece of information is redundant you discard. Typically you search for minimal systems as this tells you lots of useful things.

    Tldr yes I would regard it as a system of (n-1) equations.


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    But then if your system contains a parameter which varies the order of the system will change, which is not ideal.
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    (Original post by james22)
    But then if your system contains a parameter which varies the order of the system will change, which is not ideal.
    If you have an n-equation system depending on a vector of parameters x and for all possible x one of the equations is redundant, then it is an (n-1) equation system.

    For example we have the equations


    x+y = 1

    2(x+y)=2

    Here our parameter vector is (x,y). Clearly for all possible (x,y) the equations are equivalent so we can reduce the system.


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    (Original post by LightBlueSoldier)
    If you have an n-equation system depending on a vector of parameters x and for all possible x one of the equations is redundant, then it is an (n-1) equation system.

    For example we have the equations


    x+y = 1

    2(x+y)=2

    Here our parameter vector is (x,y). Clearly for all possible (x,y) the equations are equivalent so we can reduce the system.


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    What I meant was systems like ax+y=1, x+y=1 where a is not an unknown, but a parameter that can vary. For example it could be the weight of the car, with x and y giving it's speed and length (I realize this system is not represented by those equations, it doesn't really matter though). Then it would be a system of 2 equations, but if a=1 it is a system of 1 equation. I'm not happy with the order of the system reducing when a parameter takes certain values.
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    (Original post by james22)
    What I meant was systems like ax+y=1, x+y=1 where a is not an unknown, but a parameter that can vary. For example it could be the weight of the car, with x and y giving it's speed and length (I realize this system is not represented by those equations, it doesn't really matter though). Then it would be a system of 2 equations, but if a=1 it is a system of 1 equation. I'm not happy with the order of the system reducing when a parameter takes certain values.
    Well if you ever do linear algebra you will have to get used to it. Because that's what they do.


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    (Original post by LightBlueSoldier)
    Well if you ever do linear algebra you will have to get used to it. Because that's what they do.


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    I've just finished my second year in my maths degree, and I've done loads of linear algebra. I have never heard of what you are talking about.
 
 
 
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