Solving Problems From The Back - 2

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.

#Algebra #ProofTechniques

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$