## Disagree - Problem 2

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• disagree with your answer to the cone problem, rmilson. The answer I obtain is s=3*(1+sqrt(2))*sqrt(3-sqrt(2)) [9.1205, roughly] I believe that
Message 1 of 4238 , Jan 15, 1999
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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
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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