Prove that $ (a+b\sqrt{2})^n $ is of the form $k+l\sqrt{2}$.(a,b,k,l,n are integers; n>1) I have previously proved it for n=1. Using induction, assume $(a+b\sqrt{2})^{x-1}$ is true; it is of the form $k+l\sqrt{2}$.
for $(a+b\sqrt{2})^x$; how do i proceed from here? Binomial theorem for cases where x is even or odd?
 A: If $(a + b\sqrt{2})^{n} = k + l\sqrt{2}$ for some integers $k$ and $l$, then multiplying the equation by $a + b\sqrt{2}$, you get 
$$(a + b\sqrt{2})^{n+1} = (ak + 2bl) + (bk + al)\sqrt{2}.$$
Since $ak + 2bl$ and $bk + al$ are integers, the result is true for $n+1$. 
If you want to use the binomial theorem, a non-inductive proof may be given:
$$(a + b\sqrt{2})^{n} = \sum_{k = 0}^n \binom{n}{k}a^{n-k}b^k2^{k/2} = A + B\sqrt{2},$$
where 
$$A = \sum_{k\, \text{even}} \binom{n}{k}a^{n-k}b^k2^{k/2}\quad \text{and}\quad B =\sum_{k\, \text{odd}} \binom{n}{k}a^{n-k}b^k 2^{(k-1)/2}.$$
A: The binomial theorem would be one way through; then you don't need your own induction. Just note that every (positive) power of $b\sqrt{2}$ is either an integer or an integer times $\sqrt2$.
In order to use induction, for the inductions step you need to know that the set $\{a+b\sqrt2\mid a,b\in\mathbb Z\}$ is closed under multiplication. You can show that by explicit calculation:
$$ (a+b\sqrt2)(c+d\sqrt2) = (ac+2bd)+(ad+bc)\sqrt2 $$
