Is the following function periodic?
$$f(x)=\cos(x)*\cos(x\sqrt5)$$
A function $f$ is said to be periodic with period $P$ ($P$ being a nonzero constant) if we have $$f(x+P) = f(x)$$ for all values of $x$. If there exists a least positive constant $P$ with this property, it is called the prime period. A function with period $P$ will repeat on intervals of length $P$, and these intervals are referred to as periods.
Hint:
$$2 \cdot \cos(\alpha) \cdot \cos(\beta)= \cos(\alpha+\beta)+\cos(\alpha-\beta)$$
Hence $$ \cos(x)\cdot \cos(\sqrt{5} x) = \frac{1}{2}(\cos( (1+\sqrt{5}) x)+\cos((1-\sqrt{5}) x))$$
So in fact we are interessted if there is an $x$ such that \begin{align*} (1+\sqrt{5})x&= m \cdot 2 \pi\\ (1-\sqrt{5})x&= n\cdot 2\pi \end{align*} where $m,n\in \mathbb{N}$. As $x=0$ solves it, we only need to find another solution.
So we can say $x \neq 0$. We devide the equations as both sides are surely not equal to zero. $$ \frac{1+\sqrt{5}}{1-\sqrt{5}} = \frac{m}{n}$$ Now we simplify the left hand set
\begin{align*} \frac{1+\sqrt{5}}{1-\sqrt{5}} &=\frac{1+\sqrt{5}}{1-\sqrt{5}}\cdot \frac{1+\sqrt{5}}{1+\sqrt{5}}\\ &=\frac{(1+\sqrt{5})^2}{1-5}\\ &=-\frac{1+2\sqrt{5}+5}{4}\\ &=-\frac{3+\sqrt{5}}{2} \end{align*}
So we have
$$-\frac{3+\sqrt{5}}{2}=\frac{m}{n}$$ Now this is equal to $$-\sqrt{5}= 2\cdot \frac{m}{n}+3 $$ where the right hand side is surely in the rationals, as $m,n$ are integers and $3$ is an integer. The left hand side is not in the rationals.