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Re: PLEASE TRY, ITS A LOGICAL QUESTION

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  • guerrilla_us
    You can count backwards. When a library day falls on a Thursday, it did so for one of two reasons: 1) that was how it legitimately fell or 2) the library day
    Message 1 of 4238 , Jul 1 8:15 PM
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      You can count backwards. When a library day falls
      on a Thursday, it did so for one of two reasons: 1)
      that was how it legitimately fell or 2) the library
      day fell on Wednesday and was bumped up to Thursday.
      If you start counting backwards from Monday, you
      only encounter this once (with child the third) before
      all three children meet again, so it's not too
      hairy.<br><br>So I got Saturday, 9 days prior to their final
      Monday meeting.
    • 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, 2011
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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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