Proving ( n^4 2n^2 3n ) is Always Divisible by 6
Introduction
In this article, we explore how to prove that the polynomial ( n^4 2n^2 3n ) is always divisible by 6 for all integers ( n ). We will discuss both the use of mathematical induction and a non-inductive method that provides a comprehensive understanding of the underlying properties of the polynomial.
Mathematical Induction
To prove that ( n^4 2n^2 3n ) is divisible by 6 using mathematical induction, we follow these steps:
Base Case
First, let’s check for the base case, where ( n 1 ):
[ f(1) 1^4 2 cdot 1^2 3 cdot 1 1 2 3 6 ]
Since 6 is divisible by 6, the base case is true.
Inductive Step
Assume that the statement is true for ( k n ). That means:
[ n^4 2n^2 3n ] is divisible by 6.
Now, consider ( f(k 1) ):
[ f(k 1) (k 1)^4 2(k 1)^2 3(k 1) ]
Expanding this, we get:
[ (k 1)^4 2(k 1)^2 3(k 1) ]
[ k^4 4k^3 6k^2 4k 1 2(k^2 2k 1) 3k 3 ]
[ k^4 4k^3 6k^2 4k 1 2k^2 4k 2 3k 3 ]
[ k^4 2k^2 3k 6k^3 8k^2 7k 6 ]
[ k^4 2k^2 3k 6(k^3 k^2 k 1) ]
[ k^4 2k^2 3k 6(k^3 k^2 k 1) ]
Note that ( 6(k^3 k^2 k 1) ) is clearly divisible by 6.
To show that ( k^4 2k^2 3k ) is also divisible by 6, observe that:
[ k^4 2k^2 3k (k-1)k(k 1)(k 2) 6k ]
Since the product of any three consecutive integers is divisible by 6, and given that ( 6k ) is also divisible by 6, we can conclude that ( f(k 1) ) is divisible by 6.
Thus, by mathematical induction, ( n^4 2n^2 3n ) is divisible by 6 for all natural numbers ( n ).
Non-Inductive Proof
Alternatively, we can provide a non-inductive proof to show the same result. This proof relies on the properties of even and odd numbers and divisibility by 3.
Case 1: ( n ) is Even
If ( n ) is even, then ( n^4 ) and ( 2n^2 ) are also even, and their sum is even. Therefore, ( n^4 2n^2 ) is even. Additionally, ( 3n ) is a multiple of 3. The sum of an even number and a multiple of 3 is always a multiple of 6. Hence, ( n^4 2n^2 3n ) is divisible by 6.
Case 2: ( n ) is Odd
If ( n ) is odd, then ( n^4 ) and ( 3n ) are also odd, and their sum is even. Again, since ( 2n^2 ) is even, the polynomial ( n^4 2n^2 3n ) is even.
Next, consider the polynomial ( n^4 - 1 2n^2 - 2 3n (n^2 - 1)(n^2 1) 2n^2 - 2 3n ). Since ( n^2 - 1 ) is a product of two consecutive integers (one of which is even), it is divisible by 2. Additionally, ( n^2 1 ) and 2 are relatively prime, and thus ( (n^2 - 1)(n^2 1) ) is divisible by 4. Therefore, ( n^4 - 1 2n^2 - 2 3n ) is even.
By the properties of even numbers, we can conclude that ( n^4 2n^2 3n ) is even. Moreover, one of the numbers ( n, n 1, n 2 ) is divisible by 3. Hence, ( n^4 2n^2 3n ) is divisible by 6.