6 Greenhill Gardens,
13 / 1 / 66

My dear Sir,

Suppose á to be a prime factor of x – other than m – Then since

x^m = y^m + z^m = y + z {y + z^m-1 + mQy + z + mz^m-1}

(as is easily seen by putting (y + z - z) for y) if á divide y + z, á ^m must do so, else á would divide m or z, which it is supposed not to do.

Hence, still supposing x not m, all its factors prime or not which divide y + z do so m times. If, then, A be their product, B that of the other factors,

x + AB & y + z = A^m

y^m + z^m = x^m = A^mB^m.

Hence

A^m(B^m - 1) = (y^m - y) + (z^m - z).

Now, by a theorem of Fermat’s,

î ^m - î is always divisible by m. Hence A^m(B^m - 1) is so divisible – but A is not, by hypothesis, so B^m - 1 is; i.e. (B^m - B) + (B - 1) is divisible by m – i.e. B = mC + 1.

I know very little about the theory of numbers, having only glanced over Legendre and Barlow: and probably in consequence of defective knowledge, I have wasted much time on Fermat’s, and other, theorems. I thought I had proved that A also is of the form Cm + 1; in which case x would be of the same form – and the same reasoning would apply to one of the two y & z. The third would obviously be divisible by m. Thus it would be reduced to showing the impossibility of

(Pm + 1)^m = (Qm + 1)^m + R^mm^m.

But the proof involved a fallacy (common to such investigations) & I am not sure that the above is true.

The above gives, however, a very simple proof of the theorems you tell me are found in Barlow & Legendre.

Yours (in very great haste)
P. G. Tait.

H. F. Talbot Esq^re

[envelope:]
H. F. Talbot Esq^re
13 Great Stuart Street,
(Edin^h)

P.G. Tait