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Is (x^2 + y^2) a perfect square? (1) x is even, y is odd (2)

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Is (x^2 + y^2) a perfect square? (1) x is even, y is odd (2) [#permalink]

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16 Apr 2004, 16:56
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Is (x^2 + y^2) a perfect square?

(1) x is even, y is odd
(2) x and y are odd
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16 Apr 2004, 17:07
B it is
1) x=4 y=3 --> perfect square. x=6 y=3 not perfect square
2) no square of odd integers added together give a perfect square
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Paul

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16 Apr 2004, 18:25
Pauls statement that "no square of odd integers added together give a perfect square" was interesting.

I had to quickly prove this for my own satisfaction ... If anyones interested proof follows.

x^2 + y^2 if x, y is odd can be written as

(2n+1)^2 + (2m+1)^2
= 4n^2 + 4m^2+ 2
=> which is an even number not divisible by 4. (a)

Now we know that even perfect square must have an even root which can be represented as 2t whose square should be 4t^2
==> Which is always divisible by 4. (b)

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16 Apr 2004, 18:45
Nice work Version2. Yes, sum of perfect squares being themselves a perfect square must have a root which is divisible by 4:), Hence an even root

ie (3,4) (5,12) (12,16)
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27 May 2004, 21:12
halle. do you have the answer to this?
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04 Jan 2011, 05:05
Can I ask someone to provide a solution for this? I don't understand the solution provided by Paul. Thank you.
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04 Jan 2011, 06:37
nonameee wrote:
Can I ask someone to provide a solution for this? I don't understand the solution provided by Paul. Thank you.

Is (x^2 + y^2) a perfect square?

(1) x is even, y is odd --> clearly insufficient: if $$x=0$$ and $$y=1$$ then $$x^2+y^2=1^2$$ and the answer will YES but if $$x=2$$ and $$y=1$$ then $$x^2+y^2=5$$ and the answer will be NO. Not sufficient.

(2) x and y are odd --> $$x=2m+1$$ and $$y=2n+1$$ for some integers $$m$$ and $$n$$ --> $$x^2+y^2=(2m+1)^2+(2n+1)^2=2*(2(m^2+m+n^2+n)+1)=2*odd$$ (as $$2(m^2+m+n^2+n)+1=even+odd=odd$$), so $$x^2+y^2$$ is not a perfect square (an even perfect square must be multiple of 2^2=4, in other words $$x^2+y^2$$ to be a perfect square since it's a multiple of 2 it must be multiple of 2^2 as a perfect square has even powers of its primes, but as $$x^2+y^2=2*odd$$ then it's not a multiple of 2^2). Sufficient.

But even though formal answer to the question is B (Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient), this is not a realistic GMAT question, as: on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other, and in this question (1) says that x is even and (2) says that x is odd, so the statements clearly contradict each other.

Hope it's clear.
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Re: DS-47   [#permalink] 04 Jan 2011, 06:37
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