prove by induction: $\forall n \in N, \exists k \in N: 165^{2n} - 1 = 166k$ I am trying to prove by induction: $\forall n \in N, \exists k \in N: 165^{2n} - 1 = 166k$. 
But I've never come across having to possibly induct on two variables?
The base case for $n=1$ is true for $k = 164$
With the inductive hypothesis: assume $165^{k} - 1 = 166k$ we will prove that $165^{2k+2} - 1 = 166k$.
Where can I start on the inductive step?
 A: For $n=m\in\Bbb N$ let $165^{2m}-1 = 166 k  \implies \color{blue}{165^{2m} = 166k+1}\ , k \in \Bbb N$
For $n=m+1$,
$$165^{2m+2} - 1 = \color{blue}{165^{2m}}\cdot165^2-1 = (\color{blue}{166k+1})165^2 - 1$$
$$165^{2m+2} - 1 = 166\cdot165^2k+165^2-1 = 166\cdot165^2k+166\cdot164 = 166k'$$
$k' = 165^2k+164\in \Bbb N$
A: Hint : For induction, the base case is fine, and suppose that $165^{2n}- 1 = 166 k_n$ for some integer $k_n$ depending on $n$.
Then note that $$(165^{2(n+1)}-1) - (165^{2n}-1) = 165^{2(n+2)} - 165^{2n} = 165^{2n}(165^2-1) = 165^{2n} \times 166 \times 164$$
And find $k_{n+1}$ in terms of $k_n$ now.
A: If $f(m)=165^{2m}-1,$
$f(n+1)-165^2f(n)=-1+165^2=(165+1)(165-1)$
As $(165^2,165^2-1)=1,$
$(165^2-1)$ will divide $f(n+1)$
$\iff(165^2-1)$ divides $f(n)$
Now establish the base case
A: If you are going to use $k$ as your induction variable you will have to use a different variable your statement.  Assume $165^{2k} - 1= 166m$ and for $2k+2$ the number will change so you'll need a third variable for that.  
I suggest that you use subscripts for your $m$ variable.
We will assume $166^{2k} - 1 = 166m_k$ and we need to prove that $165^{2k+2}-1 =166m_{k+1}$ for some integer $m_{k+1}$.
Basic step:   $n=1$ then we want to prove $165^n - 1= 166m_1$ for some $m_1$.
That $165^2 -1 = (165 + 1)(165-1) = 166*164$ so if we let $m_1 = 164$ that is the base case.
Induction step.
Assume $165^{2k} -1 = 166m_k$ for some $m_k$ we want to prove that $165^{2k+2} -1 = 166*m_{k+1}$ for some integer $m_{k+1}$.
Start by noting:
$165^{2k+2}-1=165^{2k}165^2 - 1=$
$165^{2k}*165^2 - 165^2 + 165^2 - 1 =$
$165^2(165^{2k} -1) + (165^2 - 1) = $
$165^2(166m_k) + (166m_1) = $
$166[ 165^2m_k + m_1]$
So if we let $m_{k+1} = 165^2m_k + m_1$ were are done.
