We recognize the expression for $\displaystyle c$ as a perfect square trinomial which can be rewritten as:

$\displaystyle c = a^{2} + 2ab + b^{2}$

$\displaystyle c = (a + b)^{2}$

$\displaystyle c = (a + b)^{2}$ is odd if and only if the square root $\displaystyle a + b$ is odd. This happens if and only if one of $\displaystyle a$ and $\displaystyle b$ is even and one is odd - this is the correct response.

If $\displaystyle a$ and $\displaystyle b$ are both odd, then $\displaystyle ab$, the product of odd numbers, is odd, and $\displaystyle a + b + ab$, the sum of three odd numbers, is odd.

If $\displaystyle a$ and $\displaystyle b$ are both even, then $\displaystyle ab$, the product of even numbers, is even, and $\displaystyle a + b + ab$, the sum of three even numbers, is even.

If one of $\displaystyle a$ and $\displaystyle b$ is even and one is odd, then $\displaystyle ab$, which has an even factor, is even, and $\displaystyle a + b + ab$, the sum of two even numbers and an odd number, is odd.

Therefore, $\displaystyle c= a + b + ab$ is even if and only if $\displaystyle a$ and $\displaystyle b$ are both even.

For the product of two positive integers to be odd, both of the integers must themelves be odd. Therefore, if $\displaystyle ab$ is odd, it follows that both $\displaystyle a$ and $\displaystyle b$ are odd as well. We examine each of the quantities keeping this in mind.

(a) Both $\displaystyle a$ and $\displaystyle b$ are odd, so $\displaystyle a - 1$ and $\displaystyle b+1$ are even. Their product, which is $\displaystyle (a-1)(b+1)$, must be even.

(b) Since $\displaystyle ab$ is odd, both $\displaystyle ab-1$ and $\displaystyle ab+1$ are even. Their product,   $\displaystyle (ab-1)(ab+1)$, is even.

(c) Since both $\displaystyle a$ and $\displaystyle b$ are odd, their squares $\displaystyle a^{2}$ and $\displaystyle b^{2}$ are both odd; it follows that $\displaystyle a^{2}-1$ and $\displaystyle b^{2}+1$ are both even, so their product $\displaystyle (a^{2}-1)(b^{2}+1)$ is even.

The correct response is that all of the three quantites must be even.

$\displaystyle ab$ is even, so it follows that one of $\displaystyle a$ and $\displaystyle b$ is even; the other can be even or odd. Let us assume that $\displaystyle a$ will always be even; the argument will be similar if $\displaystyle b$ is assumed to always be even.

(a) $\displaystyle a$ is even, so $\displaystyle a-1$ is odd. If $\displaystyle b$ is even, then $\displaystyle b+1$ is odd, and $\displaystyle (a-1)(b+1)$, the product of odd factors, is odd. If $\displaystyle b$ is odd, then $\displaystyle b+1$ is even, and $\displaystyle (a-1)(b+1)$, having an even factor, is even. Therefore, $\displaystyle (a-1)(b+1)$ can be even or odd.

(b) $\displaystyle ab$ is even, so $\displaystyle ab -1$ and $\displaystyle ab+1$ are both odd. $\displaystyle (ab-1)(ab+1)$ is the product of odd factors and must be odd.

(c) $\displaystyle a$ is even, so $\displaystyle a^{2}$ is even as well, and $\displaystyle a^{2}-1$ is odd. If $\displaystyle b$ is even, then $\displaystyle b^{2}$ is even, and $\displaystyle b^{2}+1$  is odd; $\displaystyle (a^{2}-1)(b^{2}+1)$, the product of odd factors, is odd. If $\displaystyle b$ is odd, then $\displaystyle b^{2}$ is odd, and $\displaystyle b^{2}+1$ is even; $\displaystyle (a^{2}-1)(b^{2}+1)$, having an even factor, is even. Therefore, $\displaystyle (a^{2}-1)(b^{2}+1)$ can be even or odd.

The correct response is that only (b) need be odd.

$\displaystyle a+b$ is an odd quantity if and only if one of $\displaystyle a$ and $\displaystyle b$ is even and the other is odd. We can assume without loss of generality that $\displaystyle a$ is the even quantity and $\displaystyle b$ is the odd quantity, since a similar argument holds if $\displaystyle b$ is even.

(a) $\displaystyle b$ is odd, so $\displaystyle b+1$ is even. $\displaystyle (a-1)(b+1)$, which has an even factor, is even.

(b) $\displaystyle a$ is even and $\displaystyle b$ is odd, so $\displaystyle ab$ is even, and $\displaystyle ab -1$ and $\displaystyle ab+ 1$ are both odd. $\displaystyle (ab-1)(ab+1)$, the product of odd factors, is odd.

(c) $\displaystyle b$ is odd, so $\displaystyle b^{2}$ is odd, and $\displaystyle b^{2}+1$ is even. $\displaystyle (a^{2}-1)(b^{2}+1)$, which has an even factor, is even.

The correct response is (a) and (c) only.

Which of the following is not an integer?

The definition of an integer is, "All positive or negative whole numbers, including zero."

Therefore, our answer must pi, because pi is not a whole number. All other options fit the defintion of integers.

An integer is an positive or negative whole number, including zero. Eliminate all options which are not whole numbers and you are left with 0!

We know that $\displaystyle m$ is a positive integer with an even factors of prime numbers, for example $\displaystyle m$ could be $\displaystyle 3^{2}$ or $\displaystyle m$ could be $\displaystyle 2^{4}$, in other words, $\displaystyle m$ must be a perfect square. The only possible answer is $\displaystyle 169$, the perfect square of $\displaystyle 13$.