5. For which negative integer values of $n$ is the function $f$, defined by the equation

$$
f(x)=\cos 7 x \cdot \sin \frac{25 x}{n^{2}}
$$

a periodic function with period $T=7 \pi$?
Solution. By definition of the period, for any value of $x$ the equality must hold

$$
\cos 7(x+7 \pi) \cdot \sin \frac{25}{n^{2}}(x+7 \pi)=\cos 7 x \cdot \sin \frac{25 x}{n^{2}}
$$

Therefore, it must also hold for $x=0$. In this case, we arrive at the equation

$$
\cos 49 \pi \cdot \sin \frac{175 \pi}{n^{2}}=0
$$

Considering that $\cos 49 \pi \neq 0$ for integer negative $n$, we conclude that the equality

$$
\sin \frac{175 \pi}{n^{2}}=0
$$

holds, which is true only if $\frac{175}{n^{2}}=k$, where $k$ is an integer. Note that $175=5^{2} \cdot 7$, and therefore, among the divisors of the number 175, there are only two squares of integers: the squares of the numbers 1 and 5. However, we are only interested in the integer negative values of $n$. Thus, $n \in\{-1,-5\}$.

Substituting these values into equation (3), we verify that in both cases we obtain an identity.

Answer. $n=-1, \quad n=-5$