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If b is an integer, is sq(a^2+b^2) an integer?
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26 Feb 2011, 15:45
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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
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Re: If b is an integer, is sq(a^2+b^2) an integer?
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26 Feb 2011, 15:55



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Re: If b is an integer, is sq(a^2+b^2) an integer?
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17 Mar 2016, 16:40
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?
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17 Mar 2016, 22:05



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If b is an integer, is sqrt {a^2+b^2} an integer?
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24 Nov 2018, 08:05



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Re: If b is an integer, is sqrt {a^2+b^2} an integer?
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24 Nov 2018, 08:15
(1) Let a be \(\sqrt{3}\) and b be 2, then \(a^2 + b^2\) = 3 + 4 = 7 which is an integer, but \(\sqrt{a^2 + b^2}\)= \(\sqrt{7}\) which is not an integer. At the same time, if a = 3, b = 4, then \(a^2 + b^2\) = 25 which is an integer, but \(\sqrt{a^2 + b^2}\) = 5 which is an integer. Since we can get a Yes or No for the question, statement is Insufficient. (2) \(a^2 = 3b^2\) Substituting this, we get \(\sqrt{3b^2 + b^2}\) = \(\sqrt{4b^2}\) = 2b. Since b is an integer, 2b is also an integer. Hence Sufficient. IMO, Option B
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Re: If b is an integer, is sq(a^2+b^2) an integer?
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24 Nov 2018, 11:48
nkin wrote: (1) Let a be \(\sqrt{3}\) and b be 2, then \(a^2 + b^2\) = 3 + 4 = 7 which is an integer, but \(\sqrt{a^2 + b^2}\)= \(\sqrt{7}\) which is not an integer. At the same time, if a = 3, b = 4, then \(a^2 + b^2\) = 25 which is an integer, but \(\sqrt{a^2 + b^2}\) = 5 which is an integer.
Since we can get a Yes or No for the question, statement is Insufficient.
(2) \(a^2 = 3b^2\) Substituting this, we get \(\sqrt{3b^2 + b^2}\) = \(\sqrt{4b^2}\) = 2b. Since b is an integer, 2b is also an integer. Hence Sufficient.
IMO, Option B Hi nkin, I think didn't get when you Let a be \(\sqrt{3}\)
just double check if I understand your reasoning correctly, so when we plug in these values we get the following ? \(\sqrt{\sqrt({3})^2+2^2}\) = \(\sqrt{3+4}\) = \(\sqrt{7}\) Do i understand you correctly ? thank you



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Re: If b is an integer, is sq(a^2+b^2) an integer?
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24 Nov 2018, 11:52
Bunuel wrote: 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 > \(a^2=3b^2\) > \(\sqrt{a^2+b^2}=\sqrt{3b^2+b^2}=\sqrt{4b^2}=2b=integer\). Sufficient. Answer: B. Bunuel can you pls explain how would you get 5 ?



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Re: If b is an integer, is sq(a^2+b^2) an integer?
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24 Nov 2018, 11:58
dave13 wrote: nkin wrote: (1) Let a be \(\sqrt{3}\) and b be 2, then \(a^2 + b^2\) = 3 + 4 = 7 which is an integer, but \(\sqrt{a^2 + b^2}\)= \(\sqrt{7}\) which is not an integer. At the same time, if a = 3, b = 4, then \(a^2 + b^2\) = 25 which is an integer, but \(\sqrt{a^2 + b^2}\) = 5 which is an integer.
Since we can get a Yes or No for the question, statement is Insufficient.
(2) \(a^2 = 3b^2\) Substituting this, we get \(\sqrt{3b^2 + b^2}\) = \(\sqrt{4b^2}\) = 2b. Since b is an integer, 2b is also an integer. Hence Sufficient.
IMO, Option B Hi nkin, I think didn't get when you Let a be \(\sqrt{3}\)
just double check if I understand your reasoning correctly, so when we plug in these values we get the following ? \(\sqrt{\sqrt({3})^2+2^2}\) = \(\sqrt{3+4}\) = \(\sqrt{7}\) Do i understand you correctly ? thank you dave13, yes that is correct. Statement 1 says \(a^2 + b^2\) is an integer, so here \(a^2 + b^2\) is 7, which is an integer but sqrt(a^2 + b^2) or sqrt(7) is not an integer as asked in the question stem. This is just to give an example where using statement (1), we are not able to answer the question stem definitively.
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Re: If b is an integer, is sq(a^2+b^2) an integer?
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24 Nov 2018, 12:01
dave13 wrote: Bunuel wrote: 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 > \(a^2=3b^2\) > \(\sqrt{a^2+b^2}=\sqrt{3b^2+b^2}=\sqrt{4b^2}=2b=integer\). Sufficient. Answer: B. Bunuel can you pls explain how would you get 5 ? dave13, Here's what Bunuel is saying: If you would get 5, i.e. if a = 1 and b = 2, that a^2 + b^2 = 1 + 4 = 5, then 5 is an integer as per statement 1, but then again, sqrt (a^2 + b^2) = sqrt(5) is not an integer, as per the question stem's ask.
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Re: If b is an integer, is sq(a^2+b^2) an integer? &nbs
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