Case 1: is a multiple of either or .

If is a multiple of , it follows that , contradiction, and if is a multiple of , it follows that is not a multiple of . The same holds if is a multiple of both.

Case 2: is a multiple of neither nor .

Then for all primes dividing by Fermat's Little Theorem, and , from which it follows that the numerator is equivalent to , so the conclusion follows.

We will show that for any arbitrary prime that there always exists some such that is a multiple of , which will demonstrate that there exist no positive integers relatively prime to all the terms of the infinite sequence.

First, consider and ; taking eliminates both.

Now, any other is a multiple of neither nor . Next, taking eliminates .

I claim that if , then .

We know that where is relatively prime to by Fermat's Little Theorem. Therefore .

Let the inverses of modulo be respectively. We know that . We get that . From this it follows that . Because is a prime not equal to nor , it follows that the multiples of cycle modulo , and therefore . The conclusion follows.

my midterms are over!