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# 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 $$f(2^n) \mid f(2^{ 2^n + L f(2^n)})$$.

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.

Note by Calvin Lin
2 years, 6 months ago