6 Greenhill Gardens,

My dear Sir,

The Theorem I <enuntiated?> to you is –

If x^m = y^m + z^m

(m being a prime > 2)

x^m = α^m β^m = y^m + z^m

& α^m = y + z

if x be not divisible by m.

Thus not only is y^m + z^m/y + z (= β^m) to be an m^th power – but, the numerator and denominator must each be an m^th power.

I have also shown that β leaves a remainder 1 when divided by m.

Yours truly

P.S. Tait.

H. F. Talbot Esq^r

<envelope>

H. F. Talbot Esq^re

13 Great Stuart Street,

(Edin^r)

P.G. Tait