If n is a natural number, then (9²ⁿ − 4²ⁿ) is always divisible by -

To determine what (9^2n - 4^2n) is always divisible by for any natural number n, we can use the algebraic identity for the difference of squares.

Step-by-Step Solution
Recognize the difference of squares:
The expression (9^2n - 4^2n) can be rewritten as:
(9^n)^2 - (4^n)^2

Apply the identity a^2 - b^2 = (a - b)(a + b):
Let a = 9^n and b = 4^n. The expression becomes:
(9^n - 4^n)(9^n + 4^n)

Analyze divisibility for n = 1:
Substitute n = 1 into the expression:
9^(21) - 4^(21) = 9^2 - 4^2 = 81 - 16 = 65
The factors of 65 are 1, 5, 13, and 65.

Analyze divisibility for n = 2:
Substitute n = 2 into the expression:
9^(22) - 4^(22) = 9^4 - 4^4 = 6561 - 256 = 6305
We know 6305 / 5 = 1261 and 6305 / 13 = 485.
Since 65 (which is 5 * 13) divides both 65 and 6305, the expression is always divisible by 5 and 13.