Largest number of different values in $f(0),f(1),..,f(999)$ given $f(x)=f(398-x)=f(2158-x)=f(3214-x)$

Question

I am having trouble trying to understand the Solution (question is also linked here). The solution states that $GCD(1056, 1760) = 352$ implies that $f(x)=f(352+x)$. However we also know that $GCD(398, 2158)=2$. Wouldn't by the same logic this imply that $f(x)=f(2+x)$? It would be nice if someone can rewrite the solution or explain thoroughly . The image is from https://artofproblemsolving.com/wiki/index.php/2000_AIME_I_Problems/Problem_12

Answer 1

$f(x)=f(1056+x)$ means that $f$ is a periodic function with period dividing $1056$.
Similarly, $f(x)=f(1760+x)$ implies that the period divides $1760$.

On the other hand, $f(x)=f(398-x)$ means that $f$ is a function symmetric with respect to the vertical axis $x=199$.

The subtle difference between the plus/minus sign in the parentheses leads to the difference between periodic function and 'symmetric' function, which are different in nature. So, the similar argument does not hold for the $\gcd(398, 2158)=2$.

Answer 2

Thanks for the cool question. I think you’re confused about what exactly implies $f(x) = f(352+x)$.

This is implied by $f(x) = f(1056+x)$ from the first line of their solution, together with $f(x) = f(1760+x)$ from the second line.

Notice, however, to derive those expressions, you needed to “use” the given equation twice.

The same-logic derivation using 398 and 3214 would not be what you said; it would be $f(x)=f(2816+x)$, and $2816$ is a multiple of $352$, so this is redundant, given we already know period $\le$ 352.