Divisibility Proof for n(n²-1)(n+2) by 12 using Factorization

In summary, the conversation discusses proving the divisibility of n(n²-1)(n+2) by 12 for any n ∈ Z. A solution is attempted by assuming n = k and using induction, but the conversation ends with a request for help on how to proceed. A hint is given to factor (n²-1) as a possible next step.
  • #1
twoski
181
2

Homework Statement



Prove that for any n ∈ Z, n(n² − 1)(n + 2) is divisible by 12 .

The Attempt at a Solution



We first assume n = k for some value k.

Next we assume k(k² − 1)(k + 2) = 12m for some value m.

I don't know where to go from here. I don't think this is supposed to be an induction proof because our professor never explained induction to us yet. Every other proof I've seen for questions like this use induction (because it's so much easier to)...
 
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  • #2
hi twoski! :smile:

(try using the X2 button just above the Reply box :wink:)

hint: factor (n2 - 1) :wink:
 

FAQ: Divisibility Proof for n(n²-1)(n+2) by 12 using Factorization

How do you prove that a number is divisible by another number?

To prove that a number is divisible by another number, you can use the division algorithm or the Euclidean algorithm. The division algorithm states that for two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. If r = 0, then a is divisible by b. The Euclidean algorithm is a more efficient method that involves finding the greatest common divisor (GCD) of the two numbers and using it to determine divisibility.

What is the difference between a proof of divisibility and a proof by induction?

A proof of divisibility shows that one number is a factor of another number, while a proof by induction is used to show a statement is true for all natural numbers. In a proof of divisibility, you are showing that a certain number can be divided evenly into another number, while in a proof by induction, you are showing that a statement holds true for all natural numbers, starting from a base case and using a recursive argument.

Can you prove that a number is divisible by a certain prime number?

Yes, you can prove that a number is divisible by a certain prime number using the fundamental theorem of arithmetic. This theorem states that every positive integer can be represented as a unique product of prime numbers. Therefore, if a number is divisible by a certain prime number, it must be a factor in the prime factorization of that number.

How can you use modular arithmetic to prove divisibility?

Modular arithmetic is a useful tool in proving divisibility because it allows us to work with remainders. One way to use modular arithmetic to prove divisibility is by using the concept of congruence, where two numbers have the same remainder when divided by a certain number. If two numbers are congruent modulo a certain number, then they have the same divisibility properties.

Are there any special cases in divisibility proofs?

Yes, there are special cases in divisibility proofs, such as when a number is divisible by 2 or 5. For example, if a number ends in 0 or 5, it is divisible by 5. If a number ends in an even digit, it is divisible by 2. There are also special divisibility rules for certain numbers, such as 3, 6, 9, and 11. These rules can make the process of proving divisibility easier.

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