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# If b is an integer, is sq(a^2+b^2) an integer?

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If b is an integer, is sq(a^2+b^2) an integer? [#permalink]

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26 Feb 2011, 16:45
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If b is an integer, is $$\sqrt {a^2+b^2}$$ an integer?

(1) a^2 + b^2 is an integer.
(2) a^2 – 3b^2 = 0
[Reveal] Spoiler: OA
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Re: If b is an integer, is sq(a^2+b^2) an integer? [#permalink]

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26 Feb 2011, 16:55
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banksy wrote:
If b is an integer, is sq(a^2+b^2) an integer?
(1) a2 + b2 is an integer.
(2) a^2 – 3b^2 = 0

If b is an integer, is sqrt(a^2+b^2) an integer?

(1) a^2 + b^2 is an integer. Clearly insufficient: if a^2 + b^2 is a perfect square, say 4, then the answer is YES but if a^2 + b^2 is NOT a perfect square, say 5, then the answer is NO. Not sufficient.

(2) a^2 – 3b^2 = 0 --> $$a^2=3b^2$$ --> $$\sqrt{a^2+b^2}=\sqrt{3b^2+b^2}=\sqrt{4b^2}=2|b|=integer$$. Sufficient.

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Re: If b is an integer, is sq(a^2+b^2) an integer? [#permalink]

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17 Mar 2016, 05:15
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Re: If b is an integer, is sq(a^2+b^2) an integer? [#permalink]

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17 Mar 2016, 17:40
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Request to please format the question properly. "sq" does not imply sqrt and is misleading.

Thanks.
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Re: If b is an integer, is sq(a^2+b^2) an integer? [#permalink]

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17 Mar 2016, 23:05
vabhs192003 wrote:
Request to please format the question properly. "sq" does not imply sqrt and is misleading.

Thanks.

Edited as suggested. Thank you.
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Re: If b is an integer, is sq(a^2+b^2) an integer?   [#permalink] 17 Mar 2016, 23:05
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