Pick any prime number greater than three. Find one less than the square of that prime number. What is the greatest positive integer that must be a divisor of the result?

Source: mathcontest.olemiss.edu 10/10/2011

**SOLUTION**

Let be a prime number greater than three. The table below lists the value of for some prime numbers.

The table suggests that 24 is the greatest positive divisor of . The presence of the two consecutive even factors and gives us ideas on how to prove the conjecture.

Since , we want to show that at least three 2s and one 3 come from the product

**Idea 1**

is easy to see because both and are even and each contributes one 2.

**Idea 2**

If and are two consecutive even numbers, then 4 divides either or .

Proof

If 4 divides , we are done.

If 4 does not divide , then there exist a quotient and a remainder such that

,

If or

or

This is not possible because in either case the right-hand side is an odd number and is even.

If

The last equation shows that 4 divides . DONE.

Applying Idea 2

and are two consecutive even numbers, thus 4 divides either one of them. So we pick up an additional 2 to have .

**Idea 3**

If are three consecutive positive integers, then 3 divides one of them.

Proof

If 3 divides , we are done.

If 3 does not divide , then there exist a quotient and a remainder such that

,

If

The last equation shows that 3 divides .

If

The last equation shows that 3 divides . DONE.

Applying Idea 3

are three consecutive positive integers, thus 3 divides either or and not because it is a prime number. So we pick up a factor 3 to finally have .

Therefore, 24 is the greatest positive integer that divides .

**Answer**: Given in solution.