If x and y are integers, is x^2-y^2 odd?

[fskilnik@GMATH]

GMATH practice exercise (Quant Class 16)

If x and y are integers, is x^2-y^2 odd?

(1) x+y is odd
(2) x-y is odd

Answer: [spoiler]_____(D)___[/spoiler]

[fskilnik@GMATH]

GMATH practice exercise (Quant Class 16)

If x and y are integers, is x^2-y^2 odd?

(1) x+y is odd
(2) x-y is odd

$$x,y\,\,{\rm{ints}}\,\,\,\,\,\left( * \right)$$
$${x^2} - {y^2}\,\,\mathop = \limits^? \,\,{\rm{odd}}$$

First Approach: ("the smart way")
$$\left( 1 \right)\,\,x + y\,\,{\rm{odd}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x,y\,\,:\,\,\,{\rm{one}}\,\,{\rm{odd}}\,{\rm{,}}\,\,{\rm{another}}\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,x - y\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {**} \right)$$
$$\left( 2 \right)\,\,x - y\,\,{\rm{odd}}\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,x,y\,\,:\,\,\,{\rm{one}}\,\,{\rm{odd}}\,{\rm{,}}\,\,{\rm{another}}\,\,{\rm{even}}\,\,\,\,\, \Rightarrow \,\,\,\,x + y\,\,{\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {**} \right)$$

$$\left( {**} \right){x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) = {\rm{odd}} \cdot {\rm{odd}} = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$


Second Approach: ("the elegant way")
$$\left( 1 \right)\,\,\,x + y\,\, = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,x - y = \underbrace {x + y}_{{\rm{odd}}} - \underbrace {\,2y\,}_{\left( * \right)\,\,{\rm{even}}} = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {***} \right)$$
$$\left( 2 \right)\,\,\,x - y\,\, = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,x + y = \underbrace {x - y}_{{\rm{odd}}} + \underbrace {\,2y\,}_{\left( * \right)\,\,{\rm{even}}} = {\rm{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.\,\,\,\,\left( {***} \right)$$

$$\left( {***} \right){x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right) = {\rm{odd}} \cdot {\rm{odd}} = {\rm{odd}}\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$


The correct answer is therefore (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

[deloitte247]

$$x^2-y^2=\left(x+y\right)\left(x-y\right)$$

Statement 1
( x+ y ) is odd
If ( x - y ) is odd then
$$odd\cdot odd=odd$$
but if ( x - y ) is even then
$$odd\cdot even=even$$
Statement 1 is INSUFFICIENT.

Statement 2
If ( x + y ) is odd then
$$odd\cdot odd=odd$$
but if ( x + y ) is even , then
$$odd\cdot even=even$$
Statement 2 is INSUFFICIENT.
combining the two statement together
( x + y ) is odd and ( x - y ) is odd
product of 2 odd numbers = odd numbers
hence, product of ( x + y ) and ( x - y ) { 2 odd numbers }
$$=x^2-xy+xy-y^2$$
$$=x^2-y^2\ \ \left\{odd\ number\right\}$$
$$x^2-y^2\ is\ definitely\ an\ odd\ number$$
Both statements together are INSUFFICIENT.

$$answer\ is\ Option\ C$$

[Brent@GMATPrepNow]

If x and y are integers, is x² - y² odd?

(1) x+y is odd
(2) x-y is odd

Some important rules:
#1. ODD +/- ODD = EVEN
#2. ODD +/- EVEN = ODD
#3. EVEN +/- EVEN = EVEN

#4. (ODD)(ODD) = ODD
#5. (ODD)(EVEN) = EVEN
#6. (EVEN)(EVEN) = EVEN

Target question: Is x² - y² odd?
IMPORTANT: Notice that we can factor x² - y² to get: x² - y² = (x + y)(x - y)

Statement 1: x+y is odd
From rule #2, we can conclude that one of the values (x or y) must be ODD, and the other value must be EVEN
This means x - y must also be ODD
So, we get: x² - y² = (x + y)(x - y) = (ODD)(ODD) = ODD
So, the answer to the target question is YES, x² - y²is odd
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: x-y is odd
From rule #2, we can conclude that one of the values (x or y) must be ODD, and the other value must be EVEN
This means x + y must also be ODD
So, we get: x² - y² = (x + y)(x - y) = (ODD)(ODD) = ODD
So, the answer to the target question is YES, x² - y²is odd
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent