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 + \sqrt{2})^{n} = A_{n} + B_{n} \sqrt{2}

A_{n + 1} + B_{n + 1} \sqrt{2} = (1 + \sqrt{2}) (A_{n} + B_{n} \sqrt{2}) = A_{n} + 2 B_{n} + (A_{n} + B_{n}) \sqrt{2}

A_{n + 1}

and

A_{n}

have the same parity which means that all the

A_{i}

are odd. The values of

B

are alternately odd and even, odd when

n

is odd and even when

n

is even, that is

B_{2 i}

is even and

B_{2 i + 1}

is odd.

Method 2: By the Binomial Theorem

$(1 + \sqrt{2})^{n} = 1 + n \sqrt{2} + (\binom{n}{2}) (\sqrt{2})^{2} + (\binom{n}{3}) \sqrt{2}^{3} + . . . .$

All the Binomial coefficients are integers and so all the coefficients from the $(\binom{n}{2}) (\sqrt{2})^{2}$ onwards are even. It follows that all the terms independent of $\sqrt{2}$ have odd coefficients. The coefficients of $\sqrt{2}$ are odd when $n$ is odd and even when $n$ is even.

Generalisation:

$(1 + \sqrt{p})^{n} = 1 + n \sqrt{p} + (\binom{n}{2}) (\sqrt{p})^{2} + (\binom{n}{3}) \sqrt{p}^{3} + . . . .$

so writing

$(1 + \sqrt{p})^{n} = A_{n} + B_{n} \sqrt{p}$

it follows that $A_{n} \equiv 1$ (mod $p$ ) and always odd and $B_{n} \equiv n$ (mod $p$ ) so it is odd or even according to whether $n$ is odd or even mod $p$ .