Solving Problems From The Back - 4

How can we possibly show this? We have such little control over anything. Wait, Does showing that 2Lf(2n)2n2n(modf(2n)) 2^{ L f( 2^n)} \equiv 2^{ n - 2^n} \pmod { f(2^n)} remind us of anything? The Lefthandside makes it so tempting for us to want to apply Fermat's Little Theorem. Oh, dang! If only f(2n)f(2^n) was a prime ...

Such wishful thinking. How do we "make" it a prime? Well, if it isn't a prime, how about we take a prime factor pp. Now, we backtrack our breadcrumbs to fix it. Remember when I said that breadcrumb 2 is too strong?

Breadcrumb 2B: For all nn, for any pf(2n) p \mid f(2^n), then there exists a kk such that pf(k) p \mid f(k) and pf(2k) p \mid f(2^k).

Breadcrumb 3B: Let's classify (possible) candidates for kk. From the 1st well known result, k=2n+Lpk = 2^n + Lp, where LL is any integer, work.

Breadcrumb 4B: We want to show that there is some LL such that pf(22n+Lp) p \mid f(2^{ 2^n + L p}).

Exercise 8: What do you think Breadcrumb 5B looks like?

Ponder this, and then move on to the next note in this set.

Note by Calvin Lin
7 years ago

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