Originally Posted by skipjack

It contains various slips. For example, your reference to "any other element" is inappropriate (as you intend to show that such an element exists, you shouldn't assume it does).

Originally Posted by Prove It

My method:

The statement "24 divides $\displaystyle \begin{align*} 5^{2\,n} - 1 \end{align*}$ for all $\displaystyle \begin{align*} n \in \mathbf{N} \end{align*}$" is equivalent to "$\displaystyle \begin{align*} 5^{2\,n} - 1 = 24\,p \end{align*}$ where $\displaystyle \begin{align*} n, p \in \mathbf{N} \end{align*}$".

Base Step:

$\displaystyle \begin{align*} 5^{2 \cdot 1} - 1 &= 5^2 - 1 \\ &= 25 - 1 \\ &= 24 \\ &= 24 \cdot 1 \end{align*}$

The base step is true.

Inductive Step:

Assume that $\displaystyle \begin{align*} 5^{2\,k} - 1 = 24\,m \end{align*}$ is a true statement for some $\displaystyle \begin{align*} k, m \in \mathbf{N} \end{align*}$. Then

$\displaystyle \begin{align*} 5^{2\,\left( k + 1 \right) } - 1 &= 5^{2\,k + 2} - 1 \\ &= 5^{2\,k} \cdot 5^2 - 1 \\ &= 25 \cdot 5^{2\,k} - 1 \\ &= 24 \cdot 5^{2\,k} + 5^{2\,k} - 1 \\ &= 24 \cdot 5^{2\,k} + 24\,m \\ &= 24 \,\left( 5^{2\,k} + m \right) \\ &= 24\,p \textrm{ where } p = 5^{2\,k} + m \end{align*}$

The inductive step is true.

Thus we have proved that 24 divides $\displaystyle \begin{align*} 5^{2\,n} - 1 \end{align*}$ for all $\displaystyle \begin{align*} n \in \mathbf{N} \end{align*}$.