If a and b are positive integers, is √(a+b) an integer?

If √(a+b) is an integer then, (a+b) has to be a perfect square of a number, lets say x.
=> (a+b) = x^2
Now,
(1) says a< b+3, so substitute a=b+2 in the above eqn. (b+2+b) = x^2 which means 2(b+1) = x^2, this still doesn't tell me anything... so not sufficient.
(2)substitute a =b(b-1) in main eqn. (b^2 -b +b) = x^2 which means b^2 = x^2. because RHS is an int. so is LHS. thus sufficient ans is B

Hi All,

This question can be solved by TESTing VALUES and pattern-matching.

We're told that A and B are POSITIVE INTEGERS. We're asked if \sqrt{(A+B} is an INTEGER. This is a YES/NO question.

Fact 1: A < B+3

IF.....
A = 1
B = 3
Then \sqrt{(1+3)} = 2 and the answer to the question is YES.

IF....
A = 1
B = 1
Then \sqrt{(1+1)} = NOT an integer and the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact: 2: A = B(B-1)

IF....
B = 2
A = 2(1) = 2
Then \sqrt{(2+2)} = 2 and the answer to the question is YES.

IF....
B = 3
A = (3)(2) = 6
Then \sqrt{(6+3)} = 3 and the answer to the question is YES.

IF....
B = 4
A = (4)(3) = 12
Then \sqrt{(12+4)} = 4 and the answer to the question is YES.

As you can see, as B increases, we ALWAYS end up with a perfect square "under" the square root, so we ALWAYS end up with an integer in the end.
Fact 2 is SUFFICIENT.

Final Answer:
GMAT assassins aren't born, they're made,
Rich

VERITAS PREP OFFICIAL SOLUTION

Correct Answer: B

I. If we pick numbers for a and b it is possible to create a scenario where \(\sqrt{(a+b)}\) (a = 1 and b = 3) and a scenario where \(\sqrt{(a+b)}\) is not an integer (a = 1 and b = 4). Not Sufficient.

II. If we expand a = b(b-1), we get a = b²-b. Then we can substitute b²-b for a in the original expression and we get √(b²) - b + b or √(b²) which is equal to b. Since b is an integer, \(\sqrt{(a+b)}\) is an integer. Sufficient.

Using algebra:

1) a < b +3 --> a - b< 3 -->Or a =b, a< b. Not suff because we don't know nature of numbers

2) a = b ^2 -b
_/a + b --> _/b sq + b - b = _/b^2 = POSITIVE b as mentioned in stem

Kudos, if you like this solution

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.