You can also prove this with a direct algebraic proof:

Definiton: An integer n is said to be odd if it can be written as

*n = 2k + 1*

for some integer *k*.

Proof: Let *n* be the product of three consecutive odd numbers. Then,

`n = (2k...

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You can also prove this with a direct algebraic proof:

Definiton: An integer n is said to be odd if it can be written as

*n = 2k + 1*

for some integer *k*.

Proof: Let *n* be the product of three consecutive odd numbers. Then,

`n = (2k + 1)(2k + 3)(2k + 5)`

`= 8k^3 + 36k^2 + 46k + 15`

`= 2(4k^3 + 18k^2 + 23k + 7) + 1`

Since integers are closed under addition and multiplication,

`4k^3 + 18k^2 + 23k + 7`

is an integer. Therefore *n* is odd.

Actually the product of any odd numbers is odd. I am not sure what kind of proof you are looking for. We can use the following argument, an even number is a number divisible by 2. Since all the prime factors of the product are actually the factors of the three odd numbers, 2 is not going to be one of the factors, in other words the product is not divisible by 2.

Another way to prove it will be by contradiction.

Instead of proving p=>q, we can prove that not q=> not p

Let n be the first odd integer, n+2 will be the secon, and n+4 will be the 3rd.

Let p=n*(n+2)*(n+4)

Suppose that p is an even number, then p is divisible by 2. Which means either n divisible by 2, or n+2 divisible by 2, or n+4 divisible by 2. The last statement is false, thus p is not even.