Is x^2 - y^2 an even integer?

Question

Is  x^2 - y^2  an even integer?
(1) x-y is an even integer.
(2) x is an odd integer.

Chetan's  questions

fantaisie · Answer

x^2  -  y^2  = (x-y)*(x+y)

1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) & (x+y) is odd: (x+y)(x-y) = O*O = O
If (x-y) is odd & (x+y) is even: O*E = E
NOT SUFFICIENT

2) x is an odd integer
If x is odd, & y is odd, then:
(x-y) = O - O = E or 0
(x+y) = O + O = E or 0
(x+y)(x-y) = E * E = E

If x is odd, & y is even, then:
(x-y) = O - E = O
(x+y) = O + E = O
(x+y)(x-y) = O * O = O

NOT SUFFICIENT

1 & 2) 
We know that (x-y) is even & that x is odd.
Given those circumstances, we can conclude that y is also odd:
(x-y) = O - O = E (In the formulas above we can see, that if y is even, then (x-y) would be odd)

If x is odd and y is odd, then:
(x-y)*(x+y) = E * E = E

Answer:   C

chetan2u · Answer

Pl recheck the highlighted portion..
If x-y is already given as EVEN, how can you take that as ODD

It is a bit tricky as  Divyadisha  has mentioned below....
The following point is very important from point of view of GMAT

Solution - 
We are looking at whether x^2-y^2 is EVEN integer..

lets see the statements-

(1) x-y is an even integer. 
Most of us will correctly factorize  x^2-y^2 as (x-y)*(x+y), but will say further since x-y is an EVEN integer... (x-y)(x+y) should also be even integer...
  But NO.... are we told that x and y are integers...

a)  x = 5, y = 1 ........ YES  x^2-y^2 is an even integer.....
b)  x=\frac{25}{3} , y=\frac{1}{3} ......... .x-y =\frac{25}{3}-\frac{1}{3}= \frac{24}{3} = 8............ .But  x+y = \frac{25}{3}+\frac{1}{3} =\frac{26}{3.}.............  so x^2-y^2 = 8*\frac{26}{3}.......... NO, x^2-y^2  is not an even integer..

Different answers.... 
Insuff

(2) x is an odd integer. 
Nothing about y..
Insuff...

Combined.. 
We know x-y is an EVEN integer and x is an integer, so y will also be an integer ..
and  x^2-y^2  will be EVEN integer
Suff

C

fantaisie · Answer

Thanks for pointing out such a careless mistake! My new take on the question:

1) We know that (x - y) is even, but don't know if (x+y) is even or odd
If (x-y) even & (x+y) is odd: (x+y)(x-y) = E*O = E
If (x-y) even & (x+y) is even: E*E = E
SUFFICIENT

2) x is an odd integer

If x is odd, & y is odd, then:
(x-y) = O - O = E or 0
(x+y) = O + O = E or 0
(x+y)(x-y) = E * E = E

If x is odd, & y is even, then:
(x-y) = O - E = O
(x+y) = O + E = O
(x+y)(x-y) = O * O = O
NOT SUFFICIENT

Answer:  A

Divyadisha · Answer

x^2 - y^2 an even integer? (x-y) (x+y) is even integer? If (x-y) or (x+y) or both are odd or any one is '0' then the x^2 - y^2 will be even. (1) x-y is an even integer. x-y is even integer. It can be any +ve even integer/ -ve even integer/ 0 (x+y) can be even/odd/ 0 (x-y) (x+y) can either be even or 0 (even) Sufficient. (2) x is an odd integer y can be odd, even or 0 and x^2 - y^2 can be odd or even. Not sufficient. A is the answer Though the question is made by chetan2u , I expect it to have some hidden trick

chetan2u · Answer

You are correct Divya.... the Q is a bit tricky....
No one has caught the POINT, so I am posting the answer above..

adiagr · Answer

Question: is  x^2 - y^2  an even integer.

That is: (x-y)*(x+y) an even integer?

Which means that we need to show that  either  (x-y) is even,  or   (x+y) is even.

Statement (1)

x-y is an even integer.

Sufficient.

Statement (1) 
x is an odd integer.
 x^2  will be odd.

Then for  x^2 - y^2  to be even.

y^2  will have to odd

which mans that y will have to be odd  --> This is not given

Not sufficient.

(A) is the answer.

Kurtosis · Answer

Hi,

Answer cannot be A.

It is not stated in the question that x and y are integers.

St1: (x - y) is even --> If (x + y) is a fraction there is a possibility that (x - y)(x + y) will be odd

We need St2 to ensure that both x and y are integers.

Divyadisha · Answer

The best questions to blow the minds!! Thanks chetan2u I have to remind myself again and again not to forget fractions, -ve and 0 :beat

stonecold · Answer

One of the Best Questions on Evens-odds

Here we are not told that x and y are both integers
We need to to check if x^2-y^2 is even or not

Lets look at statements

Statement 1
x-y=even
Using test cases
x=4
y=2
x-y=2
x+y=6
so x^2-y^2 will be even
let x=4.3
y=0.3
x-y=4
and x+y=4.3
hence x^2-y^2 won't be an integer=> Neither even nor odd
Hence insufficient

Statement 2
x is odd
no clue on y 
hence not sufficient 
combining the two statements

x+y=even
and x is odd
so y=even-odd => odd
so x and y are both odd
Great
Thus x^2-y^2 will be Odd^2-Odd^2 => Odd-Odd=> Even 
Hence the answer to the Question is YES
Hence sufficient

Hence C

Kinshook · Answer

Asked: Is  x^2 - y^2  an even integer?
Is (x+y)(x-y) an even integer?

(1) x-y is an even integer.
Since it is not given that x & y are integers.
NOT SUFFICIENT

(2) x is an odd integer.
No information is provided for y.
NOT SUFFICIENT

Combining (1) & (2)
(1) x-y is an even integer.
(2) x is an odd integer.
If x is odd and (x-y) is even => y is odd
(x-y) is even and (x+y) is even
(x-y)(x+y) is even
SUFFICIENT

IMO C

Aman1012 · Answer

Hi Chetnau,

In ur solution for the option 1 you took second example of fraction that cannot be possible where u shown x+y = 26/8 or something as its already mentioned its integer and integer cannot be in fraction hence I guess you need to give better answer for this

ShankSouljaBoi · Answer

try x= 4.6 y = 0.6 that means x-y is even but x=y is not even integer.
integer*not integer = not integer. A is insufficient
B is insufficient, y not commented upon.

Combining x is odd integer and x-y is even, hence y is an odd integer. Hence, x+y must be odd + odd = odd integer .
and x-y = even integer . Hence x^2-y^2 = even integer * odd integer = even integer = Sufficient.

MHIKER · Answer

(1) Both x and y can be odd; 5-3=2 (even); then  x^2 - y^2  will be even. YES
 
 Consdieral factional numbers 
 Both x and y can be even: 3.5-1.5=2;  x^2 - y^2  will not be even. No
Insufficient.

(2) No Information about y. Insufficient.

Combining both information:
X is an odd integer from option 2 then Y will be also an odd integer. So,  x^2 - y^2  will be even.

Sufficient.

Is x^2 - y^2 an even integer?

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