Answer: For each positive integer \(n\), let \(P_n\) be the statement that \(5^{2n}-1\) is divisible by \(8\).

When \(n=1\), \(5^{2}-1=25-1=24=3\times 8\), so \(P_1\) is true.

Now assume \(P_k\) is true for some positive integer \(k\). Then for some integer \(\lambda\), \(5^{2k}-1=8\lambda\).

Consider \(5^{2k+2}-1\):

\(5^{2k+2}-1=25\cdot 5^{2k}-1=24\cdot 5^{2k}+(5^{2k}-1)\).

Since \(24=3\times 8\) and \(5^{2k}-1=8\lambda\), we get

\(5^{2k+2}-1=3\times 8\cdot 5^{2k}+8\lambda=8(3\cdot 5^{2k}+\lambda)\).

So \(5^{2k+2}-1\) is divisible by \(8\), which means \(P_{k+1}\) is true.

Therefore, by mathematical induction, \(5^{2n}-1\) is divisible by \(8\) for every positive integer \(n\).

Let \(P_n\) be the statement: \(5^{2n}-1\) is divisible by \(8\).

Base case: when \(n=1\),

\(5^2-1=25-1=24=3\times 8\),

so \(P_1\) is true.

Inductive hypothesis: assume \(P_k\) is true for some positive integer \(k\). Then

\(5^{2k}-1=8\lambda\)

for some integer \(\lambda\).

Inductive step:

\(5^{2k+2}-1=25\cdot 5^{2k}-1=24\cdot 5^{2k}+(5^{2k}-1)\).

Using the inductive hypothesis,

\(5^{2k+2}-1=24\cdot 5^{2k}+8\lambda=3\times 8\cdot 5^{2k}+8\lambda\)

\(=8(3\cdot 5^{2k}+\lambda)\).

Hence \(5^{2k+2}-1\) is divisible by \(8\), so \(P_{k+1}\) is true. Therefore \(P_k\Rightarrow P_{k+1}\).

Since \(P_1\) is true and \(P_k\Rightarrow P_{k+1}\), it follows by mathematical induction that \(5^{2n}-1\) is divisible by \(8\) for every positive integer \(n\).