I How to know if a polynomial is odd or even?

[PhysicsBoyMan]

3(2k+1)³

I have written a program which calculates the value of that polynomial with different values of k. The result is always an odd number. I am having a difficult time writing a proof that states that this polynomial always returns an odd number.

I know that (2k + 1) is the general form for odd numbers. My polynomial does have (2k+1) in it, but it is altered.

Here are two different polymorphs of the same polynomial. They are just simplified versions:

3(8k³ + 12k² + 6k + 1)

24k³ + 36k² + 18k + 3

How can I prove that this is always odd?

 

[fresh_42]

$3(2k+1)^3$ has only odd factors. No $2$ in sight.

 

[PhysicsBoyMan]

I have the answer. 3 times any odd makes an even, but the third power of that even is always odd.

 

[fresh_42]

I have the answer. 3 times any odd makes an even, but the third power of that even is always odd.

3 times odd is still odd. Even numbers can be divided by 2, odd cannot.

 

[PhysicsBoyMan]

How would you know if a difficult statement like 3(2k+1)³ can be divided by 2

 

[fresh_42]

How would you know if a difficult statement like 3(2k+1)³ can be divided by 2

It can't. $3(2k+1)^3 = 3 \cdot (2k+1) \cdot (2k+1) \cdot (2k+1)$ and every one of them is odd. There cannot be a factor $2$. $2$ is a prime so it must divide a factor if it would divide the product. This is not the case, as long as $k$ is a natural or integer number. Of course you get an even product if you allow, e.g. $k=1.5$.

 

[pwsnafu]

If you know that the general form of odd numbers is $2 n + 1$ then all you need to do is write $24k^3 + 36k^2 + 18k + 3$ in that form, i.e. find $n$. All you need to do is write the 3 as $2+1$ then factor out the twos.

 

[Mark44]

How would you know if a difficult statement like 3(2k+1)³ can be divided by 2

If you know that the general form of odd numbers is $2 n + 1$ then all you need to do is write $24k^3 + 36k^2 + 18k + 3$ in that form, i.e. find $n$. All you need to do is write the 3 as $2+1$ then factor out the twos.

It's not necessary to multiply out the (2k + 1)³ factor.

Assuming that k takes on only integer values,
2k is always even, and 2k + 1 is odd.
(2k + 1)³ is the product of three odd integers, so is itself odd.
Muliplying an odd integer by 3 results in an odd integer, which by definition is not divisible by 2.
All of my statements here can be made rigorous with very little effort.

BTW, 3(2k+1)³ is an expression, not a statement. An expression has a value; a statement is either true or false. Equations and inequalities are examples of statements.

 

[member 587159]

3(2k+1)³

I have written a program which calculates the value of that polynomial with different values of k. The result is always an odd number. I am having a difficult time writing a proof that states that this polynomial always returns an odd number.

I know that (2k + 1) is the general form for odd numbers. My polynomial does have (2k+1) in it, but it is altered.

Here are two different polymorphs of the same polynomial. They are just simplified versions:

3(8k³ + 12k² + 6k + 1)

24k³ + 36k² + 18k + 3

How can I prove that this is always odd?

3(2k+1)^3
= 24k^3 + 36k^2 + 18k + 3
= 24k^3 + 36k^2 + 18k + 2 + 1
= 2*12k^3 + 2*18k^2 + 2*9k + 2*1 + 1
= 2(12k^3 + 18k^2 + 9k + 1) + 1

Then, 12k^3 + 18k^2 + 9k + 1 is a number that depends on the value of k (It doesn't matter whether this number is odd or even, since it is multiplied by 2, and 2*odd number = even number and 2*even number is even number). Now, let s = 12k^3 + 18k^2 + 9k + 1.

Therefor, we can write 2(12k^3 + 18k^2 + 9k + 1) + 1 as 2s + 1, which is the representation of an odd number.