There are two possibilities.

1)n is not a square.

2)n is a square.

If n is not a square then it has a non-trivial proper factorization

where

.

Where

are distinct.

Thus among the factors of

:

We can find its factors

.

When

is a square there are 2 possibilities.

1)n is not a square of a prime.

2)n is a square of a prime.

If n is not a square of a prime we know that

where

because it is not a prime. Thus, it has a factorization in the form

where

are distinct and the same argument applies.

If n is a square of a prime then we have a minor problem. For example if n=4=2^2 it does not work. Which is why the initial conditions of the problem says n>4. Thus we know that

and in no other way. We know that

contains one factor. But what about other another factor of

? It turns out that when

the factor

appears among

. Thus, there is another number that has a factor of

. The reason why is because

for

thus,

and is hence among one of the factors.