To prove that $8$ divides $5^{2n} - 1$ for all natural numbers $n$ using the Principle of Mathematical Induction, we proceed as follows:

Step 1: Base Case

We start with $n = 1$ and show that the statement holds true for this initial value.

For $n = 1$, $5^{2 \cdot 1} - 1 = 5^2 - 1 = 25 - 1 = 24$ Since $8$ divides $24$, the base case is satisfied.

Step 2: Inductive Hypothesis

Assume that the statement holds for some arbitrary natural number $k$, i.e., assume that $8 \mid 5^{2k} - 1.$ This means that there exists an integer $m$ such that $5^{2k} - 1 = 8m.$

Step 3: Inductive Step

We need to show that the statement holds for $k + 1$; that is, we need to prove that $8 \mid 5^{2(k+1)} - 1.$ Starting with $5^{2(k+1)} - 1$, we expand it as follows: $5^{2(k+1)} - 1 = 5^{2k + 2} - 1 = (5^{2k})(5^2) - 1 = 25 \cdot 5^{2k} - 1.$ Now, rewrite $25 \cdot 5^{2k} - 1$ as: $25 \cdot 5^{2k} - 1 = (24 \cdot 5^{2k}) + (5^{2k} - 1).$ From the inductive hypothesis, we know that $5^{2k} - 1 = 8m$, so: $(24 \cdot 5^{2k}) + (5^{2k} - 1) = 8 \cdot 3 \cdot 5^{2k} + 8m = 8(3 \cdot 5^{2k} + m).$ This shows that $5^{2(k+1)} - 1$ is divisible by $8$, as it can be written as $8$ times an integer.

Conclusion

By the Principle of Mathematical Induction, $8$ divides $5^{2n} - 1$ for all natural numbers $n$.

Would you like further explanation or details on any part of the proof?

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1. What are the key steps in proving statements by induction?
2. Can the principle of induction be applied to non-integer sets?
3. How would the proof change if the divisor was a different number than 8?
4. Can the Principle of Induction apply to inequalities as well?
5. Why is the base case essential in mathematical induction?

Tip: In proofs by induction, clearly separate each step (base case, hypothesis, and inductive step) to improve readability and logical flow.