Period of function $x\mapsto (f(x))^2$

Question

If i know, that $f(x)$ is a periodic function with period $T$, how should i prove, that $(f(x))^2$ has period $T_1 : T_1 \le T$.

I tried to use periodic function determination: $f(x)=f(x+T)$, but it wont help.

Answer 1

Then it is obvious, because $f^2(x+T)=\{f(x+T)\}^2=\{f(x)\}^2=f^2(x)$. Hence $f^2$ repeats itself after every interval of length $T$. Thus the period of $f^2$ divides $T$. Hence $T_1\leq T$.