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.

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