Periodically problematic

Plugging in $x + 30$ for $x$ , the functional equation is equivalent to $f(x + 630) = -f(x)$ We claim that such an $f$ with fundamental period $k$ exists if and only if $k \mid 1260$ but $k \nmid 630$ .

First, assume that such an $f$ exists. Then we must have $f(x + 1260) = -f(x + 630) = -(-f(x)) = f(x)$ So $1260$ is a multiple of the period, i.e. $k \mid 1260$ . But if it were true that $k \mid 630$ , then $f(x + 630) = -f(x)$ would imply $f(x) = -f(x)$ , or $f(x) = 0$ for all $x$ . $f$ is not constant, so we must have $k \nmid 630$ .

Now suppose conversely that $k \mid 1260$ but $k \nmid 630$ , and let $f(x) = \sin\left(\frac{2\pi}{k} x\right)$ . This function has fundamental period $k$ , and we check: $f(x + 630) = \sin\left(\frac{2\pi}{k} (x + 630) \right) = \sin \left( \frac{2\pi}{k} x + \frac{1260}{k} \pi \right)$ Since k does not divide $630$ , $\frac{1260}{k}$ is odd, so the above equals $\sin \left( \frac{2\pi}{k} x + (\text{odd}) \pi \right) = \sin \left( \frac{2\pi}{k} x + \pi \right)$ $= -\sin \left( \frac{2\pi}{k} x \right) = -f(x)$ as required.

We've reduced the problem to counting the number of divisors of $1260$ which are not divisors of $630$ . $1260$ factors as $2^2 \cdot 3^2 \cdot 5 \cdot 7$ , so such divisors are 4 times any divisor of $3^2 \cdot 5 \cdot 7$ . There are $3 \cdot 2 \cdot 2 = \boxed{12}$ such divisors.

We could set $x$ to be $x+630$ to have $f_k(x+600)+f_k(x+1230)=0$ . We know that $f_k(x-30)=-f_k(x+600)$ and $f_k(x+600)=-f_k(x+1230)$ , so $f_k(x-30)=f_k(x+1230)$ , and $f_k(x)=f_k(x+1260)$ . Thus, the period must be a nonnegative factor of $1260$ . However, since the function is not constant, it is not possible for $f_k(x)=f_k(x+630)$ otherwise we would have $f_k((x+30)-30)=f_k((x+30)+600)=0$ , and the function would already be a constant $0$ . Hence the period must be a factor of $1260$ that is at the same time not a factor of $630$ .

Note that all factors of $630$ are the factors of $315$ multiplied by $1$ or $2$ , while all the factors of $1260$ are the factors of $315$ multiplied by $1$ , $2$ , or $4$ . (This is because $315$ and $2$ are relatively prime, and $630=(2)(315)$ , while $1260=(4)(315)$ .

Hence the number of factors of $1260$ that are not factors of $630$ are factors of $315$ multiplied by $4$ . Since $315=(3)(3)(5)(7)$ , $315$ has $(2+1)(1+1)(1+1)=12$ factors, so there are $\boxed{12}$ possible integers $k$ .

Substituting $x+30$ in place of $x$ , we obtain $f_{k} (x)= -f_{k} (x+630)$ . But again, $f_{k} (x+630)= -f_{k} (x+630+630) = -f_{k} (x+1260)$ This implies $f_{k} x= f_{k} (x+1260)$ for all values of $x$ , i.e $1260$ is a period of $f(x)$ .

It is now easy to see that $k \leq 1260$ . We write $1260$ as $mk+n$ , where $0 \leq n < k$ by the division algorithm. Then, note that $f_k(x)= f_k(x+1260)= f_k(x+mk+n)= f_k(x+n)$ implying $n$ is also a period of $f_k(x)$ . However, $n>0$ contradicts the minimality of $k$ . Hence we must have $n = 0$ , or equivalently $k \mid 1260$ . Since $1260$ has $36$ positive divisors, this gives rise to at most $36$ possible values for $k$ .

However, note that if $k \mid 630$ , then $630$ will also be a period of $f_k(x)$ , which is a contradiction as we have $f(x)= -f(x+630)$ for all $x$ . Note that $630$ has $24$ divisors, and $630 \mid 1260$ , which implies all divisors of $630$ will also be divisors of $1260$ . So, $24$ cases have to be rejected out of the $36$ possible ones, and our answer will be $36-24= \boxed{12}$ .

If $f(x-30) + f(x+600) = 0$ , then $f(x-30) = -f(x+600)$ , thus $f(x-30) = f(x+1230)$ . This means that the period $k$ must be a divisor of $1260$ . Thus $k$ must be of the form $2^a \times 3^b \times 5^c \times 7^d$ where $a \in \{0,1, 2\}, b \in \{0,1,2\}, c \in \{0,1\}, d \in \{0,1\}$ . But we also know that $f(x)$ isn't constant, so there is an $x$ such that $f(x+600) \ne 0$ and thus $f(x+600) \ne -f(x+600)$ and thus $f(x-30) \ne f(x+600)$ . Thus the period should not be a divisor of $630$ . So $a = 2$ . This gives us $3 \times 2 \times 2 = 12$ remaining options.

For any x we have $f(x)=-f(x+630)$ which implies $f(x)=f(x+1260)$ so the period k of f(x) should be such that k should be a factor of 1260 and not a factor of 630.

$1260=2*2*3*3*7*5$

no of factors of 1260 such that theyre not factors of 630 are $1*3*2*2=12$

We have f(x) = -f(x + 630) = f(x + 1260), So k must be a divisor of 1260 but not a divisor of 630. 630 = 2.3.3.5.7 and 1260 = 2. 630, So, we have 2.2.3 = 12 possible values of k