If b is an integer, is (a^2 + b^2)^(1/2) an integer?

Question

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

DS46402.01

Bunuel · Accepted Answer

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.

Answer: B.

nkin · Answer

(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

dave13 · Answer

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 · Answer

Bunuel  can you pls explain how would you get 5 ?

nkin · Answer

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.

nkin · Answer

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.

Basshead · Answer

If b is an integer, is  \sqrt {a^2+b^2}  an integer?

(1)  a^2 + b^2  is an integer.

Insufficient. Could or could not be an integer.

(2)   a^2 – 3b^2 = 0 
 a^2 = 3b^2

=  \sqrt {a^2+3b^2} 
=  \sqrt {4b^2} 
=  2 |b|

Since we're told b is an integer, statement 2 is sufficient.

Answer is B.

wishmasterdj · Answer

1) can be any thing
e.g. a^2 = 28, b =1 so 29^1/2 is NOT an integer
INSUFF

2) a^2=3b^2

this sqrt(a^2 + b^2) = sqrt (4b2) = 2|b|
Since b is given to be integer, hence 2 |b| is an integer. SUFF

CrackverbalGMAT · Answer

If b is an integer, is  \sqrt{a^2+b^2}  an integer?

(1)  a^2 + b^2  is an integer.

Here, we dont know whether  a^2 + b^2  is a perfect square or not. So Statement 1 alone is insufficient.

(2)  a^2 – 3b^2 = 0

a^2 = 3b^2

\sqrt{a^2+b^2}  =  \sqrt{3b^2+b^2}  =  \sqrt{4b^2}  = 2b .

Its given that b is an integer , So  \sqrt{a^2+b^2}  will be an integer. Hence sufficient.

Option B is the answer.

Thanks,
Clifin J Francis.

AndrewN · Answer

NUMBER PROPERTY CONCEPT   : This may be obvious to most people on this forum, but since no one has touched on it in the thread and the question is missed by about a quarter of the people who attempt it, I thought I would draw attention to the fact that

\sqrt{0}=0

That is, even if  a  and  b  are 0 in the above expression, √0 is still 0, an integer, and  is defined , so the answer cannot be (E).

- Andrew

jabhatta2 · Answer

Hi  AndrewN  - could you please elaborate a bit more about why you say the answer CANNOT be (E), based on the pink ?

I agree with the concept mentioned in the pink above but if a^2+b^2 was say 5...then √5 WOULD NOT GIVEN YOU an integer.

So you could get a Yes/ No, making E a contender

AndrewN · Answer

Hello,  jabhatta2 . Could this be a case in which you may have overlooked some information in the problem, namely that  b is an integer ? I agree with you that, based on Statement (1) alone, the number under the radical could be 5, or it could be some perfect square. However, as others have worked out above, when you consider Statement (2) alone, you can see that  a  can be substituted out for its equivalent expression in  b :

a^2 - 3b^2 = 0

a^2 = 3b^2

\sqrt {(3b^2)+b^2}

\sqrt {4b^2}

The 4 can be factored into two 2's and placed on the outside of the radical, leaving

2\sqrt {b^2}

Since we know that, again,   b  is an integer , the value under the radical CANNOT be 5. I was pointing out that the smallest value it can be is 0, and that √0 still leads to an integer value as an answer. Thus, (B) must be the answer (and (E), by extension, cannot).

- Andrew

bumpbot · Answer

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If b is an integer, is (a^2 + b^2)^(1/2) an integer?

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If b is an integer, is (a^2 + b^2)^(1/2) an integer?

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