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Math Expert V
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If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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Question Stats: 64% (01:48) correct 36% (02:21) wrong based on 256 sessions

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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

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GMAT 1: 630 Q49 V27 If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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Bunuel wrote:
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

$$\sqrt{x^2 + y^2}$$ 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.

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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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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

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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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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

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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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(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
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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Bunuel wrote:
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

MAGOOSH OFFICIAL SOLUTION:
Attachment: rootofsquaresum_text.PNG [ 27.53 KiB | Viewed 2244 times ]

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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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Bunuel wrote:
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

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.
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GRE 1: Q169 V154 Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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Hey Bunuel
What if y>0 is added to the question stem

Regards
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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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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.
Attachments variable approach's answer probability.jpg [ 219.74 KiB | Viewed 1498 times ]

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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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Top Contributor
Bunuel wrote:
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

(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

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Re: If x is a positive integer, is (x^2 + y^2)^1/2 an integer?  [#permalink]

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