Let x and y be integers such that x < 0 < y. If x and y satisfy all of the following condi… — GRE General Test

Let $x$ and $y$ be integers such that $x < 0 < y$ . If $x$ and $y$ satisfy all of the following conditions:

1. $(-1)^{x y + x} = -1$
2. $(-1)^{x^2 y + y} = 1$
3. $y^2 - x^2 = 19$

What is the value of $x + y$ ?

Cevap: 1

Cevap

The value of

x + y

is 1.

Condition 1 dictates that

(-1)^{x y + x} = -1

, meaning

x y + x = x(y + 1)

is odd. For the product

x(y+1)

to be odd, both

x

and

y+1

must be odd, which means

x

is odd and

y

is even. Condition 3 factors as

(y - x)(y + x) = 19

. Given

x < 0 < y

, we know

y - x > y + x

. Since 19 is prime, its unique positive factor pair requires

y - x = 19

and

y + x = 1

. Solving this system yields

y = 10

and

x = -9

, which satisfies all sign and parity constraints. Thus,

x + y = -9 + 10 = 1

Adım Adım Çözüm

Analyze parity requirements from Condition 1

x

is odd and

y

is even

Since

(-1)^{x(y+1)} = -1

, the exponent

x(y+1)

must be odd, requiring both

x

and

y+1

to be odd.

Check consistency with Condition 2

Condition 2 is satisfied

x^2 y + y = y(x^2 + 1)

is always even when

y

is even, making

(-1)^{x^2 y + y} = 1

true.

Factor difference of squares and set up system using sign rules

y - x = 19

and

y + x = 1

Since 19 is prime and

x < 0 < y

y - x > y + x > 0

, forcing the factor pair to be 19 and 1.

Solve for

x

and

y

and sum them

x = -9

y = 10

, giving

x + y = 1

Adding the system yields

2y = 20 \implies y = 10

, and substituting into

y + x = 1

yields

x = -9

Anahtar Kavram

Even-Odd Exponent Rules and Sign Properties of Integers

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