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Re: need a help!

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  • Clooneman
    I m using MS Excel in Italian (which is against the rules, I know) and I haven t done it yet. In fact, I might not bother, sorry.
    Message 1 of 4238 , Mar 1, 2002
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      I'm using MS Excel in Italian (which is against the rules, I know) and I haven't done it yet. In fact, I might not bother, sorry.
    • bqllpd
      ((c/2 + x)^2 + y^2)^(n/2) + ((c/2 - x)^2 + y^2)^(n/2) = c^n where c/2 = x = -c/2 x^2+y^2=r^2 y/r=sin(T) Parametric form x= (+or- (-c^2 * r^2
      Message 4238 of 4238 , May 1 1:50 AM
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        ((c/2 + x)^2 + y^2)^(n/2) + ((c/2 - x)^2 + y^2)^(n/2) = c^n where c/2 >= x >= -c/2
        x^2+y^2=r^2
        y/r=sin(T)

        Parametric form
        x= (+or- (-c^2 * r^2 *(sin^2(T)-1))^(1/2))/c
        y=r*sin(T)
        T=0 to 2*pi Radians (but T=0 to pi/2 is sufficient).

        I had to wait this long for the solving software to catch up.

        Show that for all n>2 and c=rational, and for all x xor y rational, then x and y are never rational.

        Doing so will give a "simple" proof of Fermat's Last Theorem.


        --- In mathforfun@yahoogroups.com, jeshields_98 wrote:
        >
        > I've just joined this club, and I have theorem
        > for<br>you. It's also been posted on Spherical Cow but here
        > seemed more appropriate, so I'll just throw it out
        > here.<br><br>Theorem:<br><br>[y^2 + (c/2 - x)^2]^(n/2) +<br>[y^2 + (c/2 +
        > x)^2]^(n/2) + c^n<br><br>Proposition 1:<br><br>If c can be
        > solved in terms of x,y,n, the result is the equation for
        > the set of curves that applies to triangles of an
        > n-dimensional Pythagorean Theorem. (including the
        > circle)<br><br>Proposition 2:<br><br>If c,y are rational, y not equal to
        > zero, then x is never rational.<br><br>I have no way of
        > knowing if proposition 2 is correct, but I HAVE derived
        > the theorem and proposition 1. I will post the link
        > to this if you would like some time in the future.
        >
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