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Let $x[n]$ be a periodic with fundamental period $N$ and $y[n] = x[kn]$ where $k \in \mathbb{N}$ and $k\ge2$. Is $y[n]$ periodic? What's the fundamental period of $y[n]$?

Here is my answer:

If $N = mk$ where $m \in \mathbb{N}$ then $y[n+m] = x[k(n + \frac{N}{k})] = x[kn+N] = x[kn] = y[n]$. So in this case the fundamental period is $\frac{N}{k}$. Otherwise simplify $\frac{N}{k}$ such that $\frac{N}{k} = \frac{m}{n}$ where $\gcd(m , n) = 1$. The fundamental period is $m$ since $y[n+m] = x[k(n+\frac{nN}{k}) ] = x[kn + nN] = x[kn] = y[n]$.

I don't know whether my solution is correct and rigorous enough. For instance how can we prove there is no smaller period?

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You are correct. The following is what you have shown.

Let $x[n]$ be a periodic sequence with period N : $$ x[n] ~=~ x[n + rN] ~~~~~,~~~\forall r,N \in Z^+ $$

Define $y[n] = x[M n]$, where $M$ is a positive integer; then

$$ y[n] = x[Mn] = x[Mn + rN] = x[M(n + \frac{rN}{M})] = y[n+K] $$

$y[n]$ will be periodic with $K$, if $\frac{r ~N}{M}$ is an integer. Hence, set $r$ to any value that makes $\frac{rN}{M}$ an integer.

If $N,M$ are already coprime, then setting $r=M$ makes $K=N$ as the period of $y[n]$. If they are not already coprime, then make them coprime $M_c,N_c$, and then set $r=M_c$ to make $K = N_c$ as the period of $y[n]$.

Make $N,M$ coprime by dividing them by their greatest common divisor $g = \text{gcd}(N,M)$ :

$$ N_c = N / g ~~~,~~~~ M_c = M / g $$

Then, for the general case, the period of $y[n]$ is:

$$ \boxed{ K = \frac{N}{ \text{gcd}(N,M) } } $$

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