Simple parity test

In the first part we have:

$a ^ 2 + a + 7$

We know that:

$a^ 2 + a = a \left (a + 1 \right)$

So $a ^ 2 + a$ is the product of two consecutive numbers and one of the terms is even. Thus, any product is even.

We know that an odd number plus an even number is always an odd number, so that the first side is always odd.

In the second part, we have:

$2b + 1$

We know from the division algorithm that $2b +1$ is always odd, because $2b$ is always even.

Now we have a product of two odd numbers

By the division algorithm we can say that all odd number can be write-down in the form:

$2k + 1, k \in \mathbb{Z}$

So we have:

$a^2 + a + 7 = 2k_1 + 1 \\ 2b+1 = 2k_2 +1 \\ ~\\ \left ( a^2 + a + 7 \right) \times \left ( 2b+1 \right) = \\ \left ( 2k_1 + 1 \right) \times \left ( 2k_2 + 1 \right) = \\ 4k_1 k_2 + 2k_1 +2k_2 + 1 = \\ 2 \left ( 2k_1k_2 + k_1 + k_2 \right) + 1 = \\ 2k_3 + 1$

So, the product is always a odd number!! :D

On the one side, 2b is always even, so (2b+1) is always odd.

On the other side, regarding a

• If a is even, then both $a^2$ and $a$ are even, so $(a^2+a+7)$ is odd.
• If a is odd, then both $a^2$ and $a$ are odd, so $(a^2+a+7)$ is odd as well.

Regardless of the parity of a and b, the given expression is always odd .