Let p and q be integers such that p < 0 < q, (-1)^p + (-1)^q = 0, and p^2 q + p is an even… — GRE General Test

Let $p$ and $q$ be integers such that $p < 0 < q$ , $(-1)^p + (-1)^q = 0$ , and $p^2 q + p$ is an even integer. Which of the following expressions must be negative?

A
$(-p)^q$
B
$(-q)^p$
C
$q^p$
$p^q$ Answer
E
$(q - p)^p$

Answer

p^q

From

(-1)^p + (-1)^q = 0

p

and

q

must have opposite parities. Factoring

p^2 q + p

into

p(pq + 1)

shows that if

p

were odd,

q

would be even, leading to an odd product

\text{odd} \times \text{odd} = \text{odd}

. Because

p^2 q + p

is given as even,

p

must be an even integer and

q

must be an odd integer. Given

p < 0

p

is a negative even integer, and

q

is a positive odd integer. The expression

p^q

represents a negative base raised to an odd exponent, which is guaranteed to be negative.

Step-by-Step Solution

Determine the parities of

p

and

q

using the equation

(-1)^p + (-1)^q = 0

One of

p

q

is even and the other is odd.

For the sum

(-1)^p + (-1)^q

to equal

0

, one term must equal

1

and the other must equal

-1

, meaning

p

and

q

have opposite parities.

Analyze the parity of

p^2 q + p = p(pq + 1)

to determine which variable is even.

p

must be even, and

q

must be odd.

p

were odd,

q

would be even, making

pq

even,

pq+1

odd, and

p(pq+1)

odd. Since

p^2 q + p

is even,

p

must be even and

q

must be odd.

Evaluate the sign of

p^q

using sign rules for exponents.

p^q < 0

p

is negative (

p < 0

) and

q

is a positive odd integer. A negative number raised to an odd power is always negative.

Key Concept

Parity and sign rules for negative bases and integer exponents

Question

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