Claim: The only solutions are \( f(x) = 1\) and \( f(x) = -1\).
Exercise 2: Show that these functions satisfy the conditions.
Now, what possible result could lead to this conclusion?
Breadcrumb 1: We want to show that \( f( 2^n) = 1\) or \(-1\) for all integers \(n\).
Exercise 3: Show that if Breadcrumb 1 is true, then the claim is true.
Breadcrumb 2: We want to show that for any integer \(n\), there exists an integer \(k\) such that \( f( 2^n) \mid f ( k)\) and \( f(2^n) \mid f(2^k)\).
Exercise 4: Show that if Breadcrumb 2 is true, then Breadcrumb 1 is true.
Ponder this, and then move on to the next note in this set.