If x and y are positive integers, is x^2 – y^2 an even integer?

Question

(1) 3 is a factor of x + y.
(2) x – y is an odd integer.

chetan2u · Answer

Hi,
x^2 - y^2 will be EVEN only if both x and y are same type of integers, either BOTH odd or BOTH even..
Also x^2 - y^2 = (x-y)(x+y)... so either of x-y or x+y is even
NOTE- if x-y is even, x+y will also be even OR both will be ODD..

lets see the statements--

(1) 3 is a factor of x + y.
we do not know if x+y has a even number factor 
say x= 6 and y = 9... ans is NO
if x= 6 and y=12... ans is YES
Insuff

(2) x – y is an odd integer.
If x-y is ODD, x+y will also be ODD, 
so our answer is NO
Suff

B

iliavko · Answer

Are you sure this is correct?..

(x+y)(x-y)= even if :

even * even = even
odd * even = even
even * odd = even

so from 1) no relevant info.

from 2) we get that x-y = odd this is only possible if:

odd - even = odd
even - odd = odd

so we get no info on x,y being odd or even.

So how can we say by using statement 2) if (x+y)(x-y) is even?

We know that (x-y) is odd, but the left (x+y) can be odd or even and we can't extract any "internal info" on x or y from inside the parenthesis.

(x+y)(x-y) is even? -> it's a question, not a statement, so we can't use it as base information to figure out if (x+y) is even or odd even knowing that (x-y) is odd, so we get E.

Am I missing something?

Bunuel , can you be so kind to take a look at this?

Thank you.

Bunuel · Answer

You are saying yourself that x - y to be odd x must be odd and y even OR x must be even and y odd. What would be x + y then?

x = odd and y = even --> x + y= odd + even = odd;
x = even and y = odd --> x + y= even + odd = odd.

Thus in any case x + y = odd.

iliavko · Answer

Thank you, Bunuel , got it. I need to get some rest for today

These odd\even questions fry my brain!

hsbinfy · Answer

answer is B

lets take 2nd statement

there are only two cases here:
E-O=O, so E+O=O
O-E=O , so O+E =O

so the expression is always odd.hence our answer

for 1st case
x+y is factor of 3 so x+y can be even or odd.not suff

Divyadisha · Answer

In what case can x^2 – y^2 be even?

Both x and y are even. then E^2-E^2= Even
or Both x and y are odd. Then O^2 - O^2= Even.

(1) 3 is a factor of x + y.
Let's say x+y = 3, where x= 2 and y = 1
In this case x^2 – y^2 is not even

If x+y= 6, where x=4 and y=2

In this case x^2 – y^2 is even

Not sufficient.

(2) x – y is an odd integer.

For x-y to be odd. either x or y is even and the other integer is odd.

In both the cases, square of one integer will be odd and the other integer will be even. And O-E or E-O will be odd.

Sufficient.

B is the answer

stonecold · Answer

Great Questions to clear the concepts on evens and odds
Given info=>
x and y are integers (this is a very important piece of info)

And we need to see whether x^2-y^2 is even or not .

Lets look at statements here
Statement 1
3 is a factor of x+y 
This statement can be proven insufficient by taking test cases pretty easily
let x+y = 3
hence x-y will be odd too
so x^y-y^2 => (x+y)*(x-y) will be odd*odd => odd
Next lets take x+y=6
so x-y will be even too
so x^y-y^2 => (x+y)*(x-y) will be even*even => even

hence Insufficient

Statement 2
Here we are given that x-y is odd
so for x-y to be odd , both x and y must have different even-odd nature
Hence x+y will be odd too.
Thus x^y-y^2 => (x+y)*(x-y) will be odd*odd => odd
Hence Sufficient

Hence B

For Quick Calculation we can memorise that 
x+y and x-y will always have the same even/odd nature

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

If x and y are positive integers, is x^2 – y^2 an even integer?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards