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If both x and y are integers, is (x^2 − y^2)^(1/2) an integer?

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If both x and y are integers, is (x^2 − y^2)^(1/2) an integer?  [#permalink]

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New post 15 May 2017, 11:26
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E

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  35% (medium)

Question Stats:

69% (01:05) correct 31% (01:13) wrong based on 62 sessions

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Re: If both x and y are integers, is (x^2 − y^2)^(1/2) an integer?  [#permalink]

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New post 15 May 2017, 21:21
St1: \(x = -y\) --> \(\sqrt{(x^2 - y^2)}\)= 0
Sufficient

St2: \(x^2 - y^2 = 8\)--> \(\sqrt{(x^2 - y^2)}\) is not an integer
\(x^2 - y^2 = 64\) --> \(\sqrt{(x^2 - y^2)}\) is an integer
Not Sufficient

Answer: A
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Re: If both x and y are integers, is (x^2 − y^2)^(1/2) an integer?  [#permalink]

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New post 15 May 2017, 21:34
Bunuel wrote:
If both x and y are integers, is \(\sqrt{x^2 − y^2}\) an integer?

(1) x + y = 0

(2) x^2 − y^2 equals the cube of an integer.



\(\sqrt{x^2 − y^2}\) = \(\sqrt{(x+y)(x-y)}\)
As per qsn., is \(\sqrt{(x+y)(x-y)}\) an integer ?

Stmt 1: x+y = 0
=> \(\sqrt{x^2 − y^2}\) = \(\sqrt{0 * (x-y)}\) = 0
Hence, sufficient.

Stmt. 2 : \(x^2 − y^2 = I^3\) [I= Integer]
=> \(\sqrt{x^2 − y^2}\) = \(\sqrt{I^3}\)
Now, I can be 8 that would give the equation an integer value and can be 9 that would lead to a non integer value.
Not sufficient.

Answer should be A.
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Re: If both x and y are integers, is (x^2 − y^2)^(1/2) an integer?  [#permalink]

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New post 30 Sep 2018, 21:07
We have both x and y are integers.
We need to find out is (x^2-y^2)^1/2 is an integer or not.

Statement 1 says, x+y=0
From the question we can deduce as
(x^2-y^2)^1/2 = ((x+y)(x-y))^1/2
From statement 1 we get
((x+y)(x-y))^1/2 = ((0)(x-y))^1/2
Implies, 0^1/2 = 0
So, yes (x^2-y^2)^1/2 = 0 (An integer)

Statement 1 is sufficient.
From Statement 2 we get that x^2-y^2 equals cube of an integer.
From this statement we can try x^2-y^2 with two of the most common cases

Case 1 , say x^2-y^2 = 27 which is 3^3
So (x^2-y^2)^1/2 = (3^3)^1/2 = 3^3/2 (Not an integer)

Case 2 , say x^2-y^2 = 64 which is 4^3
So, (x^2-y^2)^1/2 = (4^3)^1/2 = 64^1/2 = 8 (An integer)

So we have two contradictory cases so, statement 2 is insufficient.
So the correct option is A.
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Re: If both x and y are integers, is (x^2 − y^2)^(1/2) an integer?  [#permalink]

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New post 01 Oct 2018, 04:44
Bunuel wrote:
If both x and y are integers, is \(\sqrt{x^2 − y^2}\) an integer?

(1) x + y = 0

(2) x^2 − y^2 equals the cube of an integer.


Let's analyze the expression:

\(\sqrt{x^2 − y^2}\) = \(\sqrt{(x+y)(x-y)}\)

(1) x + y = 0

The expression above will always equal to zero

Sufficient.

(2) x^2 − y^2 equals the cube of an integer

Let x^2 − y^2 =0 .......................Answer is yes

Let x^2 − y^2 =1 or -1 ...............Answer is yes

Let x^2 − y^2 =27 ......................Answer is No

Insufficient

P.S.: I included different values to point put that we have quick easy values to prove yes/no answers.

Answer: A.
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Re: If both x and y are integers, is (x^2 − y^2)^(1/2) an integer? &nbs [#permalink] 01 Oct 2018, 04:44
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