FreddysFingers Wrote:SOLVE!!!!!!

r = (xp − yp)/2,
s = xyzp−2/2,
t = zp/2,

which may be combined to produce the formulae

t + r = (zp + xp − yp)/2 = xp,
t − r = (zp − xp + yp)/2 = yp.

From these formulae, it follows that

t2 − r2 = (t + r)(t − r) = xpyp = 4t2 (s/t)p.

Consider the square number

[t(p−1)/2 r]2 = tp−1r2 = tp−1 [t2 − (t2 − r2)] = tp−1 [t2 − 4t2 (s/t)p ] = t [tp − 4sp ].
Let g represent the greatest common divisor of s and t, which equals zp−2/2. Then s and t may be written in terms of smaller coprime integers d and e: s = gd and t = ge.

The number e is even and divisible by 4, because e = t/g = z2 and z is even. Since d and e are coprime, d must be odd.

Since the number
t [tp − 4sp ] = gp+1 e [ep − 4dp ]

is a square and since gp+1 and e are squares, ep − 4dp must also be a square. Since e is divisible by 4, e and ep − 4dp are squares of smaller even numbers, w and q:

e = w2,
ep − 4dp = q2.
Consider (wp + q) / 2 and (wp − q) / 2. Their sum and product are given by
(wp + q) / 2 + (wp − q) / 2 = wp,
(wp + q) / 2 * (wp − q) / 2 = (w2p − q2) / 4 = (ep − (ep − 4dp)) / 4 = dp.

Thus, any common factor in these numbers would also occur in wp and dp, and would therefore imply a common factor in w and d, and hence in e and d. Thus, since e and d are coprime, (wp + q) / 2 and (wp − q) / 2 are coprime, and since their product is a pth power, they may each individually be expressed as a pth power:

(wp + q) / 2 = up,
(wp − q) / 2 = vp,

which implies that wp = up + vp. If it could be shown that w was smaller than z, then the argument of infinite descent would imply the general case of Fermat's Last Theorem.
However, w is not smaller than z. On the contrary, w equals z:

w2 = e = t/g = z2.

FreddysFingers Wrote:SOLVE!!!!!!

r = (xp − yp)/2,
s = xyzp−2/2,
t = zp/2,

which may be combined to produce the formulae

t + r = (zp + xp − yp)/2 = xp,
t − r = (zp − xp + yp)/2 = yp.

From these formulae, it follows that

t2 − r2 = (t + r)(t − r) = xpyp = 4t2 (s/t)p.

Consider the square number

[t(p−1)/2 r]2 = tp−1r2 = tp−1 [t2 − (t2 − r2)] = tp−1 [t2 − 4t2 (s/t)p ] = t [tp − 4sp ].
Let g represent the greatest common divisor of s and t, which equals zp−2/2. Then s and t may be written in terms of smaller coprime integers d and e: s = gd and t = ge.

The number e is even and divisible by 4, because e = t/g = z2 and z is even. Since d and e are coprime, d must be odd.

Since the number
t [tp − 4sp ] = gp+1 e [ep − 4dp ]

is a square and since gp+1 and e are squares, ep − 4dp must also be a square. Since e is divisible by 4, e and ep − 4dp are squares of smaller even numbers, w and q:

e = w2,
ep − 4dp = q2.
Consider (wp + q) / 2 and (wp − q) / 2. Their sum and product are given by
(wp + q) / 2 + (wp − q) / 2 = wp,
(wp + q) / 2 * (wp − q) / 2 = (w2p − q2) / 4 = (ep − (ep − 4dp)) / 4 = dp.

Thus, any common factor in these numbers would also occur in wp and dp, and would therefore imply a common factor in w and d, and hence in e and d. Thus, since e and d are coprime, (wp + q) / 2 and (wp − q) / 2 are coprime, and since their product is a pth power, they may each individually be expressed as a pth power:

(wp + q) / 2 = up,
(wp − q) / 2 = vp,

which implies that wp = up + vp. If it could be shown that w was smaller than z, then the argument of infinite descent would imply the general case of Fermat's Last Theorem.
However, w is not smaller than z. On the contrary, w equals z:

w2 = e = t/g = z2.

FreddysFingers Wrote:Like hell I can do that!!!!! LOL
I just copied and pasted some unsolvable equation.

FreddysFingers Wrote:Oh I know. I posted the equation because we learned about it in our math class.