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If both x and y are integers, is √x^2-y^2 an integer?

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If both x and y are integers, is √x^2-y^2 an integer?  [#permalink]

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New post 30 Sep 2018, 10:08
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If both x and y are integers, is √\(x^2\)-\(y^2\) an integer?
(1) x+y=0
(2)\(x^2-y^2\)equals the cube of an integer.
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If both x and y are integers, is √x^2-y^2 an integer?  [#permalink]

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New post 30 Sep 2018, 10:33
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jalice wrote:
If both x and y are integers, is √\(x^2\)-\(y^2\) an integer?
(1) x+y=0
(2)\(x^2-y^2\)equals the cube of an integer.


S1 - x+y=0
x=-y
\(\sqrt{x^2-y^2}\) = \(\sqrt{x^2-x^2}\) = 0 = Integer
Sufficient.

S2 = \(x^2-y^2\) = \(a^3\) where a is an integer
\(\sqrt{x^2-x^2} = \sqrt{a^3}\), which may or may not be an integer.
E.g. if a=2, not integer, but if a=4, then integer.
Insufficient.
Answer A.
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Re: If both x and y are integers, is √x^2-y^2 an integer?  [#permalink]

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New post 30 Sep 2018, 20:35
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Re: If both x and y are integers, is √x^2-y^2 an integer? &nbs [#permalink] 30 Sep 2018, 20:35
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If both x and y are integers, is √x^2-y^2 an integer?

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