Can Induction be done to prove statements for integers? While studying more advanced variants of induction (like strong induction), I've been wondering about the possibility to use a variant of induction to prove some statements for all integers.
My principle is the following:


*

*Prove the statement $P$ is true for a base case $k_0 \in \mathbb{Z}$

*Prove that $P(n+1)\iff P(n)$. 
(Or $P(n) \implies (P(n+1) \land P(n-1))$


From which the statement $P$ is true for all $k \in \mathbb{Z} $
As an exemple I'll prove in this way that: $$\forall z \in \mathbb{C^*},\forall k\in\mathbb{Z}: Arg(z^k)\equiv k\cdot Arg(z)\quad[2\pi]$$
Let $z \in \mathbb C^*\quad (\mathbb C^*= \mathbb C\setminus \{0\})$ , we will Prove that $$\forall k\in\mathbb{Z}: Arg(z^k)\equiv k\cdot Arg(z)\quad[2\pi]$$


*

*For the case $k = 0,\quad z^0=1\quad (z\not=0)$, $Arg(1)\equiv 0 \quad[2\pi],\quad$hence it is true for $k=0$ 

*Let $k\in \mathbb Z$
$$\begin{align}Arg(z^{k+1})&\equiv Arg(z^k\cdot z)\quad[2\pi]\\
&\equiv Arg(z^k)+ Arg(z)\quad[2\pi]\\
\end{align} 
$$
So we can deduce that:
$$Arg(z^k)\equiv k\cdot Arg(z)\quad[2\pi]\\
\iff Arg(z^k)+ Arg(z)\equiv k \cdot Arg(z) + Arg(z)\quad[2\pi]\\
\iff Arg(z^{k+1})\equiv(k+1)\cdot Arg(z)\quad [2\pi]$$

*Conclusion: as $z$ is arbitrary, then our claim is true


Is this Principle in general valid?
if so, is my proof valid?
 A: Yes, your principle is correct and your proof is correct.
If you wish to prove your principle, that can be done simply from normal induction. First argue via normal induction that $P(n+1)\iff P(n)$ and $P(k_0)$ together proves $P(k)$ for all $k\geq k_0$. Then define $Q(n):=P(-n)$ and do the same thing to prove that  $Q(n+1)\iff Q(n)$ and $Q(k_0)$ together proves $Q(k)$ for all $k\geq k_0$. This second principle can be shown to be equivalent to the other direction that you need, completing your proof.
A: Yes, your principle is valid in general. This method of induction is known, but rarely used, mainly because in most cases it is easier to show the statement just for natural numbers (including $0$) and then use some symmetry argument or other trick to show it also holds for negative numbers.
For example, let $a\in\mathbb{R}$ and $n,m\in\mathbb{N}_0$. We can show that $a^{n+m}=a^n\cdot a^m$. Now when we introduce $a^{-n}$ as $(a^{-1})^n$, we can simply show $$a^{-n+(-m)}=a^{-(n+m)}=(a^{-1})^{n+m} =(a^{-1})^n\cdot (a^{-1})^m = a^{-n}\cdot a^{-m}$$
In the third step we used the law already established for natural numbers, so no induction over $\mathbb{Z}$ is needed.
