If x is a positive integer, is (x^2 + y^2)^1/2 an integer?

Question

If x is a positive integer, is  \sqrt{x^2 + y^2}  an integer?

(1)  \sqrt{x + y}  is an integer

(2) 8x^2 - y^2 = 0

Lucky2783 · Accepted Answer

\sqrt{x^2 + y^2}  will be an integer only when Y is an integer , why ? because square root of a fractional number cannot be an integer , and X and Y are Pythagorean pair (eg: 3,4 or 6,8 etc.) .

(1)  \sqrt{x + y}  is an integer
 this tells us that Y is an integer . NOT SUFFICIENT.

(2) 8x^2 - y^2 = 0
 if we substitute 8X^2= Y^2 in  \sqrt{x^2 + y^2}  = 3X (we cannot have -3X as X is a positive integer) . 
 SUFFICIENT.

answer B

EMPOWERgmatRichC · Answer

Hi All,

DS questions are often built around distinct patterns (even if you don't immediately recognize that a pattern is there). Often, the way to prove that there's a pattern is to TEST VALUES and pay careful attention to the results.

Here, we're told that X is a POSITIVE INTEGER. We're asked if \sqrt{(X^2 + Y^2} is an integer. This is a YES/NO question.

Fact 1: \sqrt{(X+Y)} is an integer.

IF....
X = 1
Y = 0
Then \sqrt{1} IS an integer and the answer to the question is YES.

IF....
X = 1
Y = 3
Then \sqrt{10} is NOT an integer and the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact 2: 8(X^2) - Y^2 = 0

We can manipulate this equation into...

8(X^2) = Y^2

At first glance, you might not know if there's a pattern in this information, so let's TEST a few values and see if a pattern emerges....

IF...
X = 1
Y = +-\sqrt{8}
Then \sqrt{9} IS an integer and the answer to the question is YES

IF....
X = 2
Y = +-\sqrt{32}
Then \sqrt{36} IS an integer and the answer to the question is YES

IF....
X = 3
Y = +-\sqrt{72}
Then \sqrt{81} IS an integer and the answer to the question is YES

Looking at these first 3 examples, it appears that the resulting calculation will ALWAYS be a perfect square, so the answer to the question is ALWAYS YES.
Fact 2 is SUFFICIENT

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

lipsi18 · Answer

X is +ve integer, root(X^2+Y^2) =int or not?

1. root(X+y) = integer; X+Y =integer^2; still we are not sure Y^2+X^2 will be perfect square or not.
2. 8X^2-Y^2=0; therefore Y^2=8X^2; therefore X^2+Y^2=9X^2 which is perfect square, yes sufficient

Hence answer is B

Thanks,

dream21 · Answer

(1) \sqrt{x + y} is an integer Not Sufficient

(2) 8x^2 - y^2 = 0

y^2= 8x^2

\sqrt{x^2 + y^2} = (9x^2)^1/2

Sufficient

Ans B

Bunuel · Answer

MAGOOSH OFFICIAL SOLUTION: rootofsquaresum_text.PNG

mvictor · Answer

the question basically asks:
is x^2 + y^2 a perfect square?
i don't know why, but first thing that popped into my mind = Pythagorean triplets..since there it is tested the sum of the 2 squares = a perfect square...
1. sqrt(x+y)= integer.
suppose x=1, and y=3. x+y=4, and sqrt(4)=2. so the condition is satisfied. but 1^2 + 3^2 = 10, and sqrt(10) is not an integer.
since the numbers can change greatly, it is impossible to deduct anything about x^2 + y^2.

2. 8x^2 - y^2 = 0
y^2 = 8x^2
now:
x^2 +8x^2 = 9x^2.
this under square root = 3x.

so 2 alone is sufficient.

MathRevolution · Answer

If we modify the original condition and the question, we can square both sides of sqrt(x^2+y^2)=int?. Then we get x^2+y^2=int^2?. Hence, we know x^2 and y^2 are squares of certain integers. Hence, if we look at the condition 2), from y^2=8x^2, we get x^2+y^2=x^2+8x^2=9x^2=(3x)^2. The answer is always yes and the condition is sufficient. Hence, the correct answer is B.

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AbdurRakib · Answer

(1)  \sqrt{x + y}  is an integer;  when x=4,y=-3 then  \sqrt{x^2 + y^2}   is  an integer.  When x=4,y=5 then  \sqrt{x^2 + y^2}   is not  an integer, Not Sufficient

(2)  8x^2 - y^2 =0 ,or  y^2=8x^2 ; so  \sqrt{x^2 + y^2} = \sqrt{x^2 + 8x^2} = \sqrt{9x^2} =3x,Since x is a positive integer ,Thus 3x is an integer. Sufficient

Correct Answer B

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If x is a positive integer, is (x^2 + y^2)^1/2 an integer?

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GMAT Data Sufficiency (DS) Questions

If x is a positive integer, is (x^2 + y^2)^1/2 an integer?

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