Solving Problems From The Back - 3

Breadcrumb 2 is rather strong, and we might not be able to prove it. In any case, let's carry on, and we might return back to weaken it.

Breadcrumb 3: (The only sane path that we could backtrack on) Let's classify (possible) candidates for $k$ . From the 1st well known result, $k = 2^n + Lf(2^n)$ , where $L$ is any integer, work.
Exercise 5: Prove that this family of $k$ work as claimed.

Breadcrumb 4: We want to show that there is some $L$ such that $f(2^n) \mid f(2^{ 2^n + L f(2^n)})$ .
Exercise 6: Prove that Breadcrumb 3 and 4 give Breadcrumb 2.

Breadcrumb 5: We want to show that there is some $L$ such that $f(2^n) \mid 2^{ 2^n + L f(2^n)} -2^n = 2^n ( 2^{ Lf(2^n) + 2^n - n } -1)$ .
Exercise 7: Could this be true? Why, or why not?

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}`	$\boxed{123}$