Inductive Proof for Divisibilty of Elements of a Set I have the following question: "Let $f:\mathbb{N}\rightarrow\mathbb{N}$ be defined $\forall n\in\mathbb{N}$, by $f(n)=10^{9+16n}-7$, let $A=\{m\in\mathbb{N}|m$ is divisible by $11\}$, $B=\{m\in\mathbb{N}|m$ is divisible by $17\}$ and $C=\{m\in\mathbb{N}|m$ is divisible by $137\}$.  Give a proof by mathematical induction for the following proposition: $$(f(\mathbb{N})\subseteq B)\wedge(f^{-1}(A\cup C)=\emptyset)."$$  I have been given that $10^{16}-1$ will be useful sometime during the proof as a hint, but I'm not quite sure how to get there.  I've tried multiplying my function by $10^{16}$, but I'm then just left with $$10^{9+16(n+1)}-7\cdot10^{16}.$$ I'm not quite sure what to do from there.  Help would be much appreciated.
 A: First part. Let prove $f(n) \subset B$ $\forall n \geq 0$. For $n=0$ is true. The inductive hypothesis is
$$
10^{9+16n}-7 = 0 \mod 17 \Leftrightarrow 10^{9+16(n+1)}= 7 \mod 17
$$
so
$$
10^{9+16(n+1)}-7 = 10^{9+16n}10^{16}-7 =7\cdot10^{16}-7=0\mod 17
$$
For the last equality I use again WolframAlpha. So $f(\mathbb{N})\subset B$.
Second part. For $n=0$, we have $11,137 \not\mid f(0)$. The inductive hypothesis is $11,137\not\mid f(k)$ $\forall k \leq n$. Let's see for $n+1$. By reductio ad absurdum, suppose $11$ or $137\mid f(n+1) $.


*

*Suppose $11 \mid f(n+1)$. Remember $11\not \mid f(n)$ and $10^{16} = 1 \mod 11$.
$$
10^{9+16(n+1)}-7 =  10^{9+16n}10^{16} - 7 = 10^{9+16n}-7 = f(n) = 0\mod 11
$$
wich is a contradiction.

*Suppose $137 \mid f(n+1)$. Remember $137\not \mid f(n)$ and $10^{16} = 1 \mod 137$.
$$
10^{9+16(n+1)}-7 =  10^{9+16n}10^{16} - 7 = 10^{9+16n}-7 = f(n) = 0\mod 137
$$


Edit: An alternative proof to the first part. You know that
$$
a = b \mod \phi(c) \Rightarrow d^a = d^b \mod c
$$
Since $\phi(17)=16$,
$$
10^{9+16n}-7 = 10^9-7 = 0 \mod 17$$
For the last equality I used WolframAlpha.
