Method 1: Where n=0 we see that the combination A odd, B even is possible. Where n=1 we see that the combination A odd, B odd is also possible. Now if
(1 + Ö2)n = An + BnÖ2

An+1 + Bn+1Ö2=(1+Ö2)(An + BnÖ2) = An + 2Bn + (An + Bn)Ö2
so An+1 and An have the same parity which means that all the Ai are odd. The values of B are alternately odd and even, odd when n is odd and even when n is even, that is B2i is even and B2i+1 is odd.

Method 2: By the Binomial Theorem
(1 + Ö2)n = 1 + nÖ2 + æ
ç
è 		n
2
 ö
÷
ø 		(Ö2)2 + æ
ç
è 		n
3
 ö
÷
ø 		Ö23 + ....
All the Binomial coefficients are integers and so all the coefficients from the
æ
ç
è 		n
2
 ö
÷
ø 		(Ö2)2

onwards are even. It follows that all the terms independent of Ö2 have odd coefficients. The coefficients of Ö2 are odd when n is odd and even when n is even.

Generalisation:
(1 + Öp)n = 1 + nÖp + æ
ç
è 		n
2
 ö
÷
ø 		(Öp)2 + æ
ç
è 		n
3
 ö
÷
ø 		Öp3 + ....
so writing
(1 + Öp)n = An + Bn Öp
it follows that An º 1 (mod p) and always odd and Bn º n (mod p) so it is odd or even according to whether n is odd or even mod p.