We are tasked with proving that $n^7 - n$ is divisible by 7 for any positive integer $n$. In mathematical terms, we need to show:

$n^7 - n \equiv 0 \pmod{7}$

This means $n^7 - n$ leaves a remainder of 0 when divided by 7.

Steps of the Proof:

Case 1: Use of Fermat's Little Theorem

Fermat's Little Theorem states that for any integer $n$ and a prime $p$, the following holds: $n^p \equiv n \pmod{p}$ For our case, the prime $p = 7$. Therefore, Fermat's Little Theorem gives: $n^7 \equiv n \pmod{7}$ Subtracting $n$ from both sides: $n^7 - n \equiv 0 \pmod{7}$

Thus, $n^7 - n$ is divisible by 7 for any integer $n$.

Case 2: Check for small values of $n$

Let’s verify this for small values of $n$.

• For $n = 1$: $n^7 - n = 1^7 - 1 = 0 \quad (\text{which is divisible by 7})$
• For $n = 2$: $n^7 - n = 2^7 - 2 = 128 - 2 = 126 \quad (\text{which is divisible by 7 as } 126 \div 7 = 18)$
• For $n = 3$: $n^7 - n = 3^7 - 3 = 2187 - 3 = 2184 \quad (\text{which is divisible by 7 as } 2184 \div 7 = 312)$
• For $n = 4$: $n^7 - n = 4^7 - 4 = 16384 - 4 = 16380 \quad (\text{which is divisible by 7 as } 16380 \div 7 = 2340)$

Thus, Fermat’s theorem and small cases both show that $n^7 - n$ is divisible by 7 for any integer $n$.

Conclusion:

By Fermat’s Little Theorem, $n^7 - n$ is divisible by 7 for all integers $n$. This completes the proof.

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Would you like further clarification or details?

Here are five related questions you might find interesting:

1. Can Fermat's Little Theorem be applied to prove divisibility by other primes?
2. What is the general form of Fermat’s Little Theorem?
3. How can this proof be extended for higher powers of $n$ and different moduli?
4. What is the Euler's Theorem, and how does it generalize Fermat's Little Theorem?
5. How can we prove divisibility by primes using modular arithmetic techniques?

Tip: Fermat's Little Theorem is useful for simplifying powers in modular arithmetic, especially when working with prime moduli.