Bina-ring

Thank you Andrei from Tudor Vianu National College, Bucharest, Romania for this solution.

First I looked at the values of $A$ and $B$ for the first values of $n$ in the expansion of $(1 + \sqrt 2)^n = A + B\sqrt 2$. The results I obtained are written in the following table:

$n$	$A$	$B$
1	1	1
2	3	2
3	7	5
4	17	12
5	41	29

In the cases studied, I observe that $A$ is odd for all $n$, and that $B$ is odd for $n$ odd and even for $n$ even. Now, I shall generalise this result by induction for any $n$:

First I assume that $A(n)$ and $B(n)$ are odd and I prove that $A(n+1)$ is odd and $B(n+1)$ is even. This corresponds to $n$ odd. I shall consider $A(n) = 2p+1$ and $B(n) = 2q+1$.

So, $(1 + \sqrt 2)^n = (2p+1) + (2q+1)\sqrt 2$. $$(1 + \sqrt 2)^{n+1} = (1 + \sqrt 2)^n (1+\sqrt 2) = ((2p+1) + (2q+1)\sqrt 2)(1+ \sqrt 2) = (2p+1 + 2(2q+1)) + \sqrt 2 (2q+1+2p+1).$$ So, $A(n+1) = 2(p+2q+1) + 1$ and $B(n+1) = 2(p+q+1).$ These values that I have obtained for A(n+1) and B(n+1) confirm my prediction.

In a similar manner I assume $A(n)$ odd and $B(n)$ even:
$A(n) = 2p+1$, $B(n) = 2q$. I obtain:
$A(n+1) = 2(p+2q) + 1$ and $B(n+1) = 2(p+q) + 1$.
Now, my proof by induction is complete.
The results are summarized in the following table:

	$A$ even	$A$ odd
$B$ even	-	$n$ even
$B$ odd	-	$n$ odd

Working in a similar manner for $(a + \sqrt p)^n$, I summarize my results in the following table:

	$n$	$A$	$B$
$a$ odd, $p$ odd	all $n$	even exception $n=1, A=1$	even exception $n=1, B=1$
$a$ even, $p$ even	all $n$	even	even exception $n=1, B=1$
$a$ odd, $p$ even	odd even	odd odd	odd even
$a$ even, $p$ odd	odd even	even odd	odd even