6 Greenhill Gardens,

11 / 1 / 66.

My dear Sir,

The Theorem I enuntiated <sic> to you is –

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

(m being a prime > 2)

x^m = α ^{mβ m} = y^m + z^m

& α ^m= y + z

if x be not divisible by m.

Thus not only is <maths expression>

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. G. Tait.

H. F. Talbot Esq^re

<envelope:>

H. F. Talbot Esq^re

13 Great Stuart Street,

(Edinr^h)

P. G. Tait.