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?

For $n=m\in\Bbb N$ let $165^{2m}-1 = 166 k \implies \color{blue}{165^{2m} = 166k+1}\ , k \in \Bbb N$
For $n=m+1$,

$$165^{2m+2} - 1 = \color{blue}{165^{2m}}\cdot165^2-1 = (\color{blue}{166k+1})165^2 - 1$$ $$165^{2m+2} - 1 = 166\cdot165^2k+165^2-1 = 166\cdot165^2k+166\cdot164 = 166k'$$

$k' = 165^2k+164\in \Bbb N$