Re: PLEASE TRY, ITS A LOGICAL QUESTION
- 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
- ((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= (+or- (-c^2 * r^2 *(sin^2(T)-1))^(1/2))/c
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 firstname.lastname@example.org, 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.