Question

Prove that if \(a\) is a natural number, then there exist two unequal natural numbers \(k\) and \(\ell\) for which \(a^{k}-a^{\ell}\) is divisible by 10 .

Short answer

Based on the cyclic pattern of the last digit of natural number powers, we can prove that for any natural number \(a\), there exists two unequal powers \(k=4\) and \(\ell=1\) where their difference, \(a^{k}-a^{\ell}\), is divisible by 10.

Step-by-step solution

Observation of Number Patterns

To start, notice a pattern in the last digit of natural numbers when raised to a power. This pattern repeats every 4 cycles. For example, for the number 2, the last digit of its powers are: \(2^1 = 2\), \(2^2 = 4\), \(2^3 = 8\), \(2^4 = 6\), \(2^5 = 2\), \(2^6 = 4\), \(2^7 = 8\), \(2^8 = 6\), and so on.

Steps to Prove the Statement 1

To prove the statement, take two different powers within the cycle of the repeating last digit. For powers of natural numbers, they start repeating after the 4th power. So, we select \(k=4\) (the starting point of the repetition) and \(\ell=1\) (within the cycle of repetition).

Steps to Prove the Statement 2

With these values of \(k\) and \(\ell\), we can see that the difference between \(a^4\) and \(a^1 = a\) is \(a^4 - a\). The last digit of this difference will be 0 for any natural number \(a\) (since the last digit repeats every 4 cycles), which proves that \(a^4 - a\) (when \(k=4\), \(\ell=1\)) is divisible by 10.