Proving the Divisibility of (m^5n - mn^5) by 30: A Comprehensive Approach

Introduction

The expression (m^5n - mn^5) often arises in number theory and can be shown to be divisible by 30. This article explores various methods to prove this statement, leveraging both modular arithmetic and prime factorization techniques. By demonstrating divisibility by the prime factors of 30 (i.e., 2, 3, and 5), we can assert the overall divisibility by 30.

Using Modular Arithmetic

A straightforward method to prove that (m^5n - mn^5) is divisible by 30 involves modular arithmetic. We examine the expression in the context of its modulo 2, 3, and 5.

Modulo 2

Let's consider the expression modulo 2:

Expanding (m^5n - mn^5), we get: (m^5n - mn^5 m^2n - n^2)

Since (m^2n - n^2 (m-n)(m n)), we can rewrite the expression as: ((m-n)(m n))

For any integers (m) and (n), one of (m-n) or (m n) must be even, meaning that ((m-n)(m n)) is divisible by 2.

Modulo 3

Now, let's consider the expression modulo 3:

Expanding (m^5n - mn^5) using (a^2 equiv a pmod{3}) for any integer (a) not divisible by 3, we get:

(m^5n - mn^5 mnn - nmm n(m^2n - m^2n) 0 pmod{3})

This shows that (m^5n - mn^5) is divisible by 3.

Modulo 5

Finally, for modulo 5:

The expression can be simplified as:

(m^5n - mn^5 n(m^4 - n^4))

Using Fermat's Little Theorem, we know that (a^4 equiv 1 pmod{5}) for any integer (a) not divisible by 5. This implies:

(m^4 - n^4 1 - 1 0 pmod{5})

Hence, (m^5n - mn^5) is divisible by 5.

Factoring and Prime Factorization

A more detailed factoring approach can be used to confirm divisibility by 30.

Define (N mnm - mn^2n^2).

For cases where neither (m) nor (n) is divisible by 2, (mn) is divisible by 2 since the product of two odd numbers is odd.

Similarly, if neither (m) nor (n) is divisible by 3, and given certain conditions, (mn) can be shown to be divisible by 3.

For no divisors of 5, (m^2n^2) is shown to be divisible by 5 as per the rules from modular arithmetic.

Since 2, 3, and 5 are the prime factors of 30, and each of the prime factors of 30 divides a factor of (N), (N) is divisible by 30.

Conclusion

In summary, by considering the expression modulo 2, 3, and 5, and using factorization and number theory principles, we can confidently state that (m^5n - mn^5) is divisible by 30. This proof combines modular arithmetic and prime factorization to provide a comprehensive approach to solving such problems.