## Number Theory Problem

**Problem:** Define by

Prove that if is prime then has no solution for except for

**Proof:** We need a quick lemma.

**Lemma:** Suppose that where , then

**Proof:** Let denote the set of divisors of and respectively. For notational convenience let

and

Note then that if then by the assumption that we have that for some and some . But, not only is it true that if that but in fact every number of that form also divides . It follows that

So that

From where the conclusion follows.

Noting then that if for some prime we have that

So that

We can combine this with the above lemma to conclude that if

that

So, getting to the actual problem. If then and . Thus, we may assume WLOG that .

So, suppose that

then by previous comment

But, since is prime and the RHS of the above equation is a factorization of it it follows that

for some . But, expanding the RHS and subtracting one gives

and thus and thus for some . But, remembering that is prime it follows that . Thus, dividing both sides of by gives

Now, since we see that from where it follows that and thus we see that the LHS of is even whereas the RHS is odd, contradiction. The conclusion follows.

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