Analyzing the Divisibility of \(5^{2n} - 1\)

The question asks for the minimum number of natural numbers by which the expression \(5^{2n} - 1\) is always divisible for any natural number \(n\). A natural number starts from 1 (1, 2, 3, ...).

Evaluating the Expression for Specific Natural Numbers

Let's substitute a few small natural numbers for \(n\) and see what values we get:

• If \(n = 1\), the expression is \(5^{2 \times 1} - 1 = 5^2 - 1 = 25 - 1 = 24\).
• If \(n = 2\), the expression is \(5^{2 \times 2} - 1 = 5^4 - 1 = 625 - 1 = 624\).
• If \(n = 3\), the expression is \(5^{2 \times 3} - 1 = 5^6 - 1 = 15625 - 1 = 15624\).

We are looking for numbers that divide 24, 624, 15624, and any other value of \(5^{2n} - 1\) for all natural numbers \(n\). Such numbers must be common divisors of all these values. The set of numbers that always divide the expression is the set of divisors of the greatest common divisor (GCD) of the values \(5^{2n} - 1\) for \(n=1, 2, 3, \dots\). Let's find the GCD of the first two values: GCD(24, 624).

We can see that \(624 = 24 \times 26\). So, the GCD(24, 624) is 24.

Let's check if 15624 is also divisible by 24. \(15624 \div 24 = 651\). Yes, it is.

This suggests that the expression \(5^{2n} - 1\) might always be divisible by 24. Let's prove this algebraically.

Algebraic Proof of Divisibility

The expression is \(5^{2n} - 1\). We can rewrite \(5^{2n}\) as \((5^2)^n\). So the expression becomes \((5^2)^n - 1\), which is \(25^n - 1\).

We know a standard algebraic factorization: \(a^n - b^n\) is always divisible by \((a-b)\) for any natural number \(n\).

Using this property, with \(a=25\) and \(b=1\), the expression \(25^n - 1^n\) is always divisible by \((25 - 1)\), which is 24.

So, \(5^{2n} - 1\) is always divisible by 24 for any natural number \(n\).

This means that any number that divides 24 will also divide \(5^{2n} - 1\) for all natural numbers \(n\). The question asks for the minimum number of natural numbers that always divide the expression. This is equivalent to finding the number of natural divisors of 24, as these are the guaranteed common divisors for all values of \(n\).

Finding the Natural Divisors of 24

Let's list the natural numbers that divide 24 without leaving a remainder:

• 1 (since \(24 \div 1 = 24\))
• 2 (since \(24 \div 2 = 12\))
• 3 (since \(24 \div 3 = 8\))
• 4 (since \(24 \div 4 = 6\))
• 6 (since \(24 \div 6 = 4\))
• 8 (since \(24 \div 8 = 3\))
• 12 (since \(24 \div 12 = 2\))
• 24 (since \(24 \div 24 = 1\))

The natural divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.

Counting the Divisors

Counting the numbers in the list above, we find there are 8 natural divisors of 24.

Since \(5^{2n} - 1\) is always divisible by 24, it is guaranteed to be divisible by all the natural divisors of 24. There are 8 such divisors. Therefore, the expression \(5^{2n} - 1\) is always divisible by a minimum of 8 natural numbers.

The common natural divisors across all values of \(n\) are precisely the natural divisors of GCD(24, 624, 15624, ...), which we found to be 24. The number of these common divisors is the number of divisors of 24, which is 8.

Revision Table: Key Concepts for Divisibility

Additional Information: Understanding Number of Divisors

The number of divisors of a natural number can be found using its prime factorization. If a natural number \(N\) has a prime factorization given by \(N = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k}\), where \(p_1, p_2, \dots, p_k\) are distinct prime numbers and \(a_1, a_2, \dots, a_k\) are their positive integer exponents, then the total number of natural divisors of \(N\) is given by the formula:

\((a_1 + 1)(a_2 + 1)\dots(a_k + 1)\)

In our case, the number 24 has the prime factorization \(24 = 8 \times 3 = 2^3 \times 3^1\).

Here, \(p_1=2\) with exponent \(a_1=3\), and \(p_2=3\) with exponent \(a_2=1\).

Using the formula, the number of divisors of 24 is \((3 + 1)(1 + 1) = 4 \times 2 = 8\).

This confirms our count of the divisors of 24.