((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.

>