Induction step other than $1$

Question

Let $k \in \mathbb{N}$. I'm trying to prove that the following implication is true only when $k=1$: $$[\exists n_0 \ \ \ P(n_0)] \land[\forall n\ge n_0 \ \ \ P(n) \implies P(n+k)] \implies [\forall n\ge n_0 \ \ \ P(n) ]$$ Obviously in the case $k=1$ we have the axiom of induction. How can we construct counterexamples for the other cases $k\not = 1$? Is it possible to give only one counterexample which works for all $k\not = 1$?

Answer 1

Fix your $k, n_0$ and let $P(n)$ be true iff $n \equiv n_0 \pmod{k}$. Then $P(n) \implies P(n+k)$, but $P(n)$ is only true for all $n \geq n_0$ if $k=1$.