Proving the Divisibility of n^5 - n and n^9 - n^5 by 10

In this article, we will delve into the mathematical proofs of two interesting expressions, (n^5 - n) and (n^9 - n^5), to show that both are divisible by 10 for specific cases. We will demonstrate the divisibility by 2 and 5 separately, given that 10 is the product of these two prime factors.

Divisibility of (n^5 - n) by 10

First, let's explore the simpler expression, (n^5 - n). We aim to prove that it is divisible by 10 for any integer (n).

Divisibility by 2

To check the divisibility by 2, we factorize the expression:

(n^5 - n n(n^4 - 1) n(n^2 - 1)(n^2 1))

Since one of (n) or (n^2 - 1) is always even, the entire expression is even. This proves that (n^5 - n) is divisible by 2.

Divisibility by 5

Next, we consider the expression modulo 5. The possible values of (n) modulo 5 are 0, 1, 2, 3, and 4. We will evaluate each case:

(text{if } n equiv 0 pmod{5}): (n^5 - n equiv 0 - 0 equiv 0 pmod{5}) (text{if } n equiv 1 pmod{5}): (n^5 - n equiv 1 - 1 equiv 0 pmod{5}) (text{if } n equiv 2 pmod{5}): (n^5 equiv 2^5 equiv 32 equiv 2 pmod{5}) and (n^5 - n equiv 2 - 2 equiv 0 pmod{5}) (text{if } n equiv 3 pmod{5}): (n^5 equiv 3^5 equiv 243 equiv 3 pmod{5}) and (n^5 - n equiv 3 - 3 equiv 0 pmod{5}) (text{if } n equiv 4 pmod{5}): (n^5 equiv 4^5 equiv 1024 equiv 4 pmod{5}) and (n^5 - n equiv 4 - 4 equiv 0 pmod{5})

We have shown that (n^5 - n) is divisible by 5 in all cases. Combining this with the divisibility by 2, we conclude that (n^5 - n) is divisible by 10 for all integers (n).

Divisibility of (n^9 - n^5) by 10

Next, we will prove that (n^9 - n^5) is divisible by 10 for specific cases of (n).

Divisibility by 2

First, we factorize the expression:

(n^9 - n^5 n^5(n^4 - 1) n^5(n^2 - 1)(n^2 1))

Similar to the previous expression, one of (n) or (n^2 - 1) is always even, making the entire expression even. This proves that (n^9 - n^5) is divisible by 2.

Divisibility by 5

To check divisibility by 5, we again consider the expression modulo 5. The possible values of (n) modulo 5 are 0, 1, 2, 3, and 4. We evaluate each case:

(text{if } n equiv 0 pmod{5}): (n^9 - n^5 equiv 0 - 0 equiv 0 pmod{5}) (text{if } n equiv 1 pmod{5}): (n^9 - n^5 equiv 1 - 1 equiv 0 pmod{5}) (text{if } n equiv 2 pmod{5}): (n^9 equiv 2^9 equiv 512 equiv 2 pmod{5}) and (n^5 equiv 2^5 equiv 32 equiv 2 pmod{5}), so (n^9 - n^5 equiv 2 - 2 equiv 0 pmod{5}) (text{if } n equiv 3 pmod{5}): (n^9 equiv 3^9 equiv 19683 equiv 3 pmod{5}) and (n^5 equiv 3^5 equiv 243 equiv 3 pmod{5}), so (n^9 - n^5 equiv 3 - 3 equiv 0 pmod{5}) (text{if } n equiv 4 pmod{5}): (n^9 equiv 4^9 equiv 262144 equiv 4 pmod{5}) and (n^5 equiv 4^5 equiv 1024 equiv 4 pmod{5}), so (n^9 - n^5 equiv 4 - 4 equiv 0 pmod{5})

We have shown that (n^9 - n^5) is divisible by 5 in all cases. Combining this with the divisibility by 2, we conclude that (n^9 - n^5) is divisible by 10 for (n equiv 0, 1, 2 pmod{5}). However, it is not universally divisible by 10 for all integers (n), especially for those congruent to 3 or 4 modulo 5.

Conclusion

We have proven that (n^5 - n) and (n^9 - n^5) are divisible by 10 for specific cases. For (n^5 - n), it is divisible by 10 for all integers (n). For (n^9 - n^5), it is divisible by 10 for (n equiv 0, 1, 2 pmod{5}), but not for (n equiv 3, 4 pmod{5}).

In summary, the divisibility of (n^5 - n) and (n^9 - n^5) by 10 relies on the properties of divisibility by 2 and 5, and the specific modular cases of (n).