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.

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