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

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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
nkin
(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 (a^2 + b^2)^(1/2) an integer? [#permalink]
Bunuel
banksy
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.

Bunuel can you pls explain how would you get 5 ?
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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
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dave13
nkin
(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 (a^2 + b^2)^(1/2) an integer? [#permalink]
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dave13
Bunuel
banksy
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.

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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If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
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.

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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
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
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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
Top Contributor
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.

Thanks,
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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
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banksy
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
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
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If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
AndrewN
banksy
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
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

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
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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
jabhatta2
AndrewN
banksy
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
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

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
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
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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
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Re: If b is an integer, is (a^2 + b^2)^(1/2) an integer? [#permalink]
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