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^{m}B^{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^{m}m^{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__