## Prime Power

#### Problem

Given that $p$ is prime it can be seen that $p + 1$ is a perfect cube when $p$ = 7. What is most surprising is that this is the only value of $p$ for which $p + 1$ is a perfect cube.

Prove that there only ever exists one prime value $p$ for which $p + 1$ is a perfect power and determine the condition for this perfect power to exist.

#### Solution

Let $p + 1 = a^k$.

$\therefore p = a^k - 1 = (a - 1)(a^{k-1} + a^{k-2} + ... + a^2 + a + 1)$

For RHS to be prime, $a - 1 = 1 \implies a = 2$.

Hence for $p + 1$ to be a perfect power it must be of the form $2^k$. Moreoever, as $p = 2^k - 1$ it can be seen that there will only exist a perfect $k$^{th} power if $2^k - 1$ is prime.

For example, when $k = 4$, $2^4 - 1 = 15 = 3 \times 5$, hence there will not exist a prime value of $p$ for which $p$ + 1 is a perfect fourth power.

In fact, it turns out that $p + 1$ can only be a perfect $k$^{th} power if the value of $k$ is itself prime.

Suppose that $k$ is composite; let $k = xy$.

$\therefore 2^{xy} - 1 = \left(2^x - 1\right)\left(2^{(x-1)y} + 2^{(x-2)y} + ... + 2^{2y} + 2^y + 1\right)$

But if $2^{xy} - 1$ is prime then $2^x - 1 = 1 \implies x = 1 \implies k = y$.

Similarly if $y$ had factors $x'$ and $y'$ we could show that $x' = 1$, and so on *ad infinitum*. In other words, $y = k$, which must be prime.

Does it follow that $p + 1$ can always be written as a perfect $k$^{th} power if $k$ is prime?