Excuse me if the notation below is wrong in some way, I do not have any special background and learn Math alone.
What i'm investigating now is:
Given a function $f(x)$ prove that it is periodic if $ \{ \exists\; T\ne 0 : \forall x \in D(f), \; x+T \in D(f), \; x-T \in D(f) \; \text{and} \ P \; \text{is true}\implies f(x-T) = f(x+T) \}$ where $P$ is one of the following statements: $$ f(x+T) = -f(x) \label{eq:1} \\ f(x+T) = {1\over{f(x)}} \\ f(x+T) = {{f(x) + a}\over{b\cdot f(x)-1}}\\ f(x+T) = {1 \over {1-f(x)}} $$
Case 1:
$$ f(x+T) = -f(x) \iff f(x) = -f(x-T) \\ f(x) = -f(x-T) \iff -f(x) = f(x-T) $$
Therefore $f(x) = f(x) \iff f(x+T) = f(x-T)$, and its period is $2T$
Case 2:
$$ f(x+T) = {1\over f(x)} \iff f(x) = {1\over f(x-T)} \\ {{1\over f(x)} = {1 \over f(x)}} \iff f(x+T) = f(x-T) $$
Case 3: $$ f(x+T) = {{f(x)+ a}\over b\cdot f(x)-1} \iff f(x)={{f(x-T)+a} \over {b\cdot f(x-T) -1}} \\ f(x)(b\cdot f(x-T)-1) = f(x-T) + a \\ f(x)(b\cdot f(x-T)-1) - f(x-T) = a \\ b\cdot f(x) f(x-T) - f(x) - f(x-T) = a \\ b\cdot f(x) f(x-T) - f(x-T) = f(x) + a \\ f(x-T) (b\cdot f(x) - 1) = f(x) + a \\ f(x-T) = { {f(x) + a} \over {b\cdot f(x) - 1} }\\ f(x-T) = f(x+T) $$
Case 4:
I couldn't untangle this one, what steps should I take to prove condition from case 4 makes a function periodic? I'm also interested whether the solution for the first 3 ones even makes sense.
For case 4, write out $f(x+3T)$ in terms of $f(x)$; \begin{eqnarray*} f(x+3T)&=&\frac{1}{1-f(x+2T)} =\frac{1}{1-\frac{1}{1-f(x+T)}} =\frac{1}{1-\frac{1}{1-\frac{1}{1-f(x)}}}\\ &=&\frac{1}{1-\frac{1-f(x)}{1-f(x)-1}} =\frac{1}{1+\frac{1-f(x)}{f(x)}} =\frac{1}{1+\frac{1}{f(x)}-1}=\frac{1}{\frac{1}{f(x)}}=f(x). \end{eqnarray*}
For the other three cases your reasoning is fine, but be careful with claims about a 'minimal period'. Generally all you can say is that the function $f$ is periodic with a period $2T$. It might also be periodic with other periods (for example period $T$, or $f$ could be constant). For case 3 an intermediate step for how you found the expression for $f(x-T)$ might also be appropriate.