# Solving Problems From The Back - 4

How can we possibly show this? We have such little control over anything. Wait, Does showing that $2^{ L f( 2^n)} \equiv 2^{ n - 2^n} \pmod { f(2^n)}$ remind us of anything? The Lefthandside makes it so tempting for us to want to apply Fermat's Little Theorem. Oh, dang! If only $f(2^n)$ 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 $p$. Now, we backtrack our breadcrumbs to fix it. Remember when I said that breadcrumb 2 is too strong?

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

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

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

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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