Proving that a square of an odd integer is also odd

[hackedagainanda]

Prove that for any arbitrary odd x, that x^2 is also odd.

By definition an odd number is an integer that can be written in the form of 2k + 1 for some integer k. This means that x = 2k + 1 where k is an integer

So let x^2 = (2k + 1)^2 we then get 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1, This is where I get lost, I understand that k is an integer but how does it follow that 2k^2 + 2k is an integer?

 

[mfb]

The product of an integer and an integer is an integer.
The sum of an integer and an integer is an integer.

More formally, integers are closed under addition and multiplication.

 

[hackedagainanda]

Thanks! I totally overlooked that

 

[fresh_42]

Prove that for any arbitrary odd x, that x^2 is also odd.

By definition an odd number is an integer that can be written in the form of 2k + 1 for some integer k. This means that x = 2k + 1 where k is an integer

So let x^2 = (2k + 1)^2 we then get 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1, This is where I get lost, I understand that k is an integer but how does it follow that 2k^2 + 2k is an integer?

Another way is to use the fact that $2$ is a prime and then apply the definition of a prime.