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

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    Taken from the Legendre's three square theorem page on wikipedia.

    In mathematics, Legendre's three-square theorem states that a natural number can be represented as the sum of three squares of integers

    n = x^2 + y^2 + z^2

    if and only if n is not of the form n = 4^a(8b + 7) for integers a and b.

    The first numbers that cannot be expressed as the sum of three squares (i.e. numbers that can be expressed as n = 4^a(8b + 7)) are

    7, 15, 23, 28, 31, 39, 47, 55, 60, 63, 71

    As trolls go I'd say this was a fairly sophisticated one. However it has the failing that anyone that is willing to attempt this question is more likely to know about the Legendre's three square theorem.

    So I give points for intelligence, points for sophisticatedness of the trap but take away points for mis-judging your targets.

    All in all I give this a 8/10. Wear it with pride.
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    (Original post by RichE)
    Aren't they both contractible and so homotopy equivalent to a point?
    Along with the fact that homotopy equivalence is a genuine equivalence relation, that would certainly do it. I was half-expecting the OP to read what a homotopy was and get some clumsy argument involving weird functions, if he was going to commit to his genius self-title!
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    (Original post by JAIYEKO)
    Integers, sorry
    There are no integer solutions these are the only solutions... unless you can provides evidence to the contrary

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    (Original post by Aph)
    There are no integer solutions these are the only solutions... unless you can provides evidence to the contrary

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    We have been given 3 equations for those 3 unknown parameters. There can only be one solution. The solution gives 3 numbers which are not integers. Therefore it's contradiction to say that the equations hold for integers.
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    (Original post by JAIYEKO)

    The closest person yet
    By closest yet you either mean:

    (a) I spotted the typo in your question;
    (b) You posted a question with no solutions
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    The question has no solutions for integers. Many of us have already showed this one way or another. (unless it's a typo, but the OP denied that)
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    (Original post by Spairos)
    We have been given 3 equations for those 3 unknown parameters. There can only be one solution. The solution gives 3 numbers which are not integers. Therefore it's contradiction to say that the equations hold for integers.
    Are you sure about that?
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    (Original post by BabyMaths)
    Are you sure about that?
    Isn't it a system of equations or do we treat each equation individually?
    I consider that there is at most one solution given the stated equations. (the one I typed earlier, given that there are no inegers constraints and treating the parameters as exchangeable ones)
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    (Original post by JAIYEKO)
    I think i know what i wrote
    There are no integer solutions.
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    (Original post by Spairos)
    We have been given 3 equations for those 3 unknown parameters. There can only be one solution. The solution gives 3 numbers which are not integers. Therefore it's contradiction to say that the equations hold for integers.
    There can be multiple solutions. Consider the simpler system

    x+y=1
    2x+2y=2
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    (Original post by Spairos)
    Isn't it a system of equations or do we treat each equation individually?
    I consider that there is at most one solution given the stated equations. (the one I typed earlier, given that there are no inegers constraints)
    If (a,b,c) is a solution then so are (a,c,b), (b,a,c) etc.
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    (Original post by james22)
    There can be multiple solutions. Consider the simpler system

    x+y=1
    2x+2y=2
    But I am talking about the given equations of this question. The second equation of yours is the same as the first one.
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    (Original post by BabyMaths)
    If (a,b,c) is a solution then so are (a,c,b), (b,a,c) etc.


    I treat them as exchangeable parameters
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    (Original post by JAIYEKO)
    I think i know what i wrote
    Nobody else does...
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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.
    It was a simpler example. There may be 1 solution to the original problem, but you cannot say that just by saying that there are 3 equations in 3 unknowns.
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    (Original post by Spairos)
    Isn't it a system of equations or do we treat each equation individually?
    I consider that there is at most one solution given the stated equations. (the one I typed earlier, given that there are no inegers constraints and treating the parameters as exchangeable ones)
    There are 6 solutions in this case, though in some sense they're essentially the same.

    If a,b,c are the three numbers you wrote earlier then

    (a,b,c), (b,c,a), (c,a,b), (c,b,a), (b,a,c), (a,c,b)

    are all solutions.

    You may think this a bit of a cop-out so it's worth noting that three equations in three variables can easily lead to no, finitely many or infinitely many solutions.
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    (Original post by RichE)
    There are 6 solutions in this case, though in some sense they're essentially the same.

    If a,b,c are the three numbers you wrote earlier then

    (a,b,c), (b,c,a), (c,a,b), (c,b,a), (b,a,c), (a,c,b)

    are all solutions.

    You may think this a bit of a cop-out so it's worth noting that three equations in three variables can easily lead to no, finitely many or infinitely many solutions.
    Indeed, you are right, I didn't express clearly what I wanted to say (that I was talking about a solution of a set which doesn't have integers and hence there are no integers as solutions because the set doesn't include any and was specifically talking about this system which doesn't have more solutions than this set - that's why I typed 3 values which I didn't set them to be equal to x, y, z).
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    Scrub-level troll. gtfo
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    (Original post by james22)
    There can be multiple solutions. Consider the simpler system

    x+y=1
    2x+2y=2
    But that is only 1 equation...
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    (Original post by james22)
    There can be multiple solutions. Consider the simpler system

    x+y=1
    2x+2y=2
    We are not in primary school-.
 
 
 
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