A friend of mine recently posed a question for me to solve after I had mentioned to him that I had been wanting to get back into mathematics: Can you show that 3 is the only prime number followed by a perfect square? A prime number being an integer with only two factors, one and itself. A perfect square being an integer \(s\) that is equal to another integer \(z\) multiplied by itself.

To answer this question it must be shown that the following does not hold:

Let \(p \in \mathbb{P}\), \(p > 3\)

Let \(z \in \mathbb{N}\)

\(p + 1 = z ^ 2\)

Proof:

Rewrite the equation

\(p + 1 = z^2\)

\(p = z^2 - 1\)

\(p = (z + 1)(z - 1)\)

Show that \(z + 1 > 1\) when \(z > 2\)

This is trivial, as any integer greater than 2 added to one will be greater than one.

Show that \(z - 1 > 1\) when \(z > 2\)

If z = 3, then \(z - 1 = 2\) and \(2 > 1\).

Since neither \((z + 1)\) or \((z - 1)\) can be \(1\) when \(z > 2\), this violates the definition of a prime number. A prime number only has two factors: itself and the number \(1\).

This shows that \(\forall p \in \mathbb{P} > 3\), the following number \(p + 1\) is not a perfect square.

QED