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hardest maths question, genius whoever solves it?

The function f from the real numbers to the real numbers satisfies f(x+y)= (1+yx+1)f(x)+(1+xy+1)f(y)+x2y+xy+xy2. If f(1)=4, what is the value of f(8)?

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  • Anonymous
    9 years ago
    Favorite Answer

    f(x + y) = (1 + xy + 1)f(y) + (x^2)y + xy + x(y^2)

    f(x + y) = (xy + 2)f(y) + xy(x + 1 + y)

    f(x + y) = xy*f(y) + 2*f(y) + xy(x + y + 1)

    f(x + y) = xy(x + y + 1 + f(y)) + 2*f(y)

    Let u = x + y, xy = y(u - y)

    f(u) = y(u - y)(u + 1 + f(y)) + 2*f(y)

    f(1) = 4, so substitute u = 1:

    4 = y(1 - y)(2 + f(y)) + 2*f(y)

    (y - y^2)(2 + f(y)) + 2*f(y) - 4 = 0

    2y - 2y^2 + y*f(y) - (y^2)*f(y) + 2*f(y) - 4 = 0

    We need to find a formula for f(y) such that this equation will always hold true for all values for y:

    f(y)*(y^2 - y - 2) = (2y - y^2 - 4)

    f(y) = (2y - y^2 - 4) / (y^2 - y - 2)

    Therefore:

    f(u) = y(u - y)(u + 1 + (2y - y^2 - 4) / (y^2 - y - 2)) + 2*(2y - y^2 - 4) / (y^2 - y - 2)

    Hence f(8) = y(8 - y)(9 + (2y - y^2 - 4) / (y^2 - y - 2)) + 2*(2y - y^2 - 4) / (y^2 - y - 2))

    More information is needed to find a numerical answer.

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