1730.31 – Counter-Factual Theory

Exactly one of the following is true for every prime number p>2p > 2. Which one is it?

  1. (p−1)(p−12)−1(p-1)^{\small\left(\dfrac{p-1}{2}\right)} - 1 is evenly divisible by p−2p-2,

  2. (p−1)(p−12)(p-1)^{\small\left(\dfrac{p-1}{2}\right)} is evenly divisible by pp,

  3. (p−1)(p−12)+1(p-1)^{\small\left(\dfrac{p-1}{2}\right)} + 1 is evenly divisible by p+1p+1,

  4. (p−1)(p−12)−1(p-1)^{\small\left(\dfrac{p-1}{2}\right)} - 1 is evenly divisible by p−1p-1.

Solution

For this problem, it’s much easier to discover which of the given answers fail. All but the first, as it happens, can be disproven by a single counterexample such as this one: consider p=5p = 5. Then

(p−1)(p−12)=(5−1)(5−12)=42=16\begin{align*}(p-1)^{\small\left(\dfrac{p - 1}{2}\right)} &= (5 - 1)^{\small\left(\dfrac{5 - 1}{2}\right)} \\[0.5em]&= 4^2 \\[0.5em]&= 16\end{align*}

Case b is violated: sixteen is not evenly divisible by five.

Case c is violated: seventeen is not evenly divisible by six.

Case d is violated, fifteen is not evenly divisible by four.

However, a is satisfied: fifteen is evenly divisible by three. Because we have been told that one of these four cases is always true, and we have found that three of them fail at least once, we know that a is the one that is always true. (Supposedly.)