Lemma 1: are positive integer, .

Proof: The if part is , for the only if part: let where are integers and . Since , hence . Since , hence .


Original problem: We will first show that by induction:

Base case:
Inductive step: assume that the statement is true for all up to , we will show that it's true for . Indeed: and by Lemma 1 which is true by assumption.


Next we will show with induction again that if an odd prime satisfies , then :

We go straight to the step: Assuming that , we will show

. By the result above for all . By a well known lemma: (note that since ), therefore .


Finally we are going to show where is an odd prime and together using induction.

Base case: because . .

Inductive Step: Assuming that the statement is true up to and all odd primes(not 3) less than , we proceed in two steps:

Step 1: If is a prime, then we have to show . Since by our assumption that and it's well known that , hence and we are done.

Step 2: Whether is a prime or not, we will show by first proving all odd prime powers but in, denoted by , divide , in other words , which is equivalent to . Since is obvious, we just have to worry about

By our assumption: , we will prove , hence giving . We have to prove which is true shown in*.

Now we will deal with primes individually:

is straight forward: .

For , as we've seen, it doesn't divide but , hence we have . We want to prove which is true for and we are done.

*To show that is true, for convinience let or which is true for our scenario.