Proving the Divisibility of ( n^5 - 5n^3 - 4n ) by 120 for ( n geq 2 )

In this article, we will explore how to prove that the expression ( n^5 - 5n^3 - 4n ) is divisible by 120 for all integers ( n geq 2 ). This involves understanding the prime factorization of 120 and factoring the given expression to check its divisibility by 8, 3, and 5. Let's delve into the detailed steps.

Step 1: Factor the Expression

First, let's factor the expression ( n^5 - 5n^3 - 4n ). We start by factoring out the common term ( n ), obtaining:

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

Next, we notice that ( n^4 - 5n^2 - 4 ) can be treated as a quadratic in terms of ( n^2 ). Let ( x n^2 ), and the expression becomes:

( x^2 - 5x - 4 )

Factoring this quadratic, we get:

( (x - 4)(x 1) )

Substituting back ( x n^2 ), we find:

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

Further factoring ( n^2 - 4 ) as a difference of squares:

( ((n - 2)(n 2))(n^2 1) )

Thus, the original expression can be rewritten as:

( n(n - 2)(n 2)(n - 1)(n 1) )

Step 2: Check Divisibility by 8

To check for divisibility by 8, we need to consider the factors of the product ( n(n - 2)(n 2)(n - 1)(n 1) ).

Case 1: ( n ) is Even

If ( n ) is even, then ( n ) contributes at least two factors of 2. Additionally, one of ( n - 2 ) or ( n 2 ) must also be even, contributing another factor of 2. Therefore, the product has at least three factors of 2, ensuring divisibility by 8.

Case 2: ( n ) is Odd

If ( n ) is odd, the consecutive even numbers in the product are ( n - 1 ) and ( n 1 ). Two consecutive even numbers provide at least four factors of 2, ensuring divisibility by 8.

Step 3: Check Divisibility by 3

Among the five consecutive integers ( n, n - 2, n 2, n - 1, n 1 ), at least one of them must be divisible by 3, since every set of three consecutive integers contains a multiple of 3. Therefore, the product is divisible by 3.

Step 4: Check Divisibility by 5

Similarly, among the five consecutive integers ( n, n - 2, n 2, n - 1, n 1 ), at least one of them must be divisible by 5, since every set of five consecutive integers contains a multiple of 5. Therefore, the product is divisible by 5.

Conclusion

Since we have shown that ( n(n - 2)(n 2)(n - 1)(n 1) ) is divisible by 8, 3, and 5, we can conclude that:

( n^5 - 5n^3 - 4n ) is divisible by 120 for ( n geq 2 ).

Additionally, it's worth noting that ( n(n - 2)(n 2)(n - 1)(n 1) ) is the product of five consecutive integers. One of any five consecutive integers must be a multiple of 5, one of any three consecutive integers must be a multiple of 3, and since we have more than three consecutive integers, there must be at least one multiple of 3 in there. We also have at least two even numbers: if ( n ) is even, then ( n - 2 ) and ( n 2 ) are even; if ( n ) is odd, then ( n - 1 ) and ( n 1 ) are even. One of any two consecutive even numbers is a multiple of 4, ensuring that the product is a multiple of 8. Since the product is a multiple of 5, 3, and 8, it must also be a multiple of ( 5 times 3 times 8 120 ).

It might also be interesting to point out that ( n(n - 2)(n 2)(n - 1)(n 1) ) is the permutation of ( n^2 ) and 5, emphasizing the elegant nature of the expression.