If m and n are positive integers, is root (n-m) an integer?

m,n are positive integers and we need to find whether \(\sqrt{n-m} = integer\).

Per statement 1, n>m+15.

Take 2 cases:

case 1, when n=m+16, then you get "yes" for \(\sqrt{n-m} = integer\) but
case 2, when n=m+17, then you get "no" for \(\sqrt{n-m} = integer\)

Thus this statement is not sufficient.

Per statement 2, n=m(m+1) --->\(\sqrt{n-m} = \sqrt{m^2} = m\)=integer. Sufficient.

B is the correct answer.
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Is \(\sqrt{n - m}\) an integer, given that \(m\) and \(n\) are positive integers?

Statement 1

\(n > m + 15\)

Easiest way to say that this statement is insufficient would be by thinking of it as statement that says, \(n > m\)
This just means that \(\sqrt{n - m}\) is not an irrational number. This is insufficient information to answer the question

Another point of view is as follows
Assume that \(\sqrt{n - m}\) is a positive integer greater than another positive integer, say x.

\(\sqrt{n - m} > x\)
squaring both sides
\(n - m = x^2\)
\(n = m + x^2\)

This equation is similar to statement 1, however, one may be tempted to say that \(\sqrt{15}\) is not an integer, hence, the whole argument falls apart and \(\sqrt{n - m}\) is not an integer. But that's not the case.

If \(n > m + 15\)
then, \(n\) is also greater than \(m + 9\)
and \(\sqrt{9}\) is an integer.

Also, if \(n > m + 15\)
then, \(n\) is also greater than \(m + 4\) and \(n\) is also greater than \(m + 1\)
and \(\sqrt{4}\), \(\sqrt{1}\)are integers.

Two different answers, hence the statement is insufficient.


Statement 2

\(n = m(m+1)\)
\(n = m^2+m\)....... (#1)

Substituting (#1) in \(\sqrt{n - m}\)

\(\sqrt{m^2 + m - m}\)
\(\sqrt{m^2}\)
\(m\)

m is an integer, as stated in the question, hence, \(\sqrt{n - m}\) is an integer.

Statement 2 is sufficient.


Answer: B
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This is a Yes-No DS question. Start by using plugging in for such questions.

m,n --> Positive integers Is √(n−m) an integer?

1) n > m + 15
n - m > 15

n - m = 16 Answer to the question [Is √(n−m) an integer?] -- Yes
n - m = 17 Answer to the question [Is √(n−m) an integer?] -- No

Insufficient

2) n = m(m+1)

m = 1, n = 2 => √(n−m) = 1 Answer to the question [Is √(n−m) an integer?] -- Yes
m = 2, n = 6 => √(n−m) = 2 Answer to the question [Is √(n−m) an integer?] -- Yes
m = 3, n = 12 => √(n−m) = 3 Answer to the question [Is √(n−m) an integer?] -- Yes
m = 6, n = 42 => √(n−m) = 6 Answer to the question [Is √(n−m) an integer?] -- Yes

Sufficient

Thus, the answer to the question is B.

Hope that helped! Point to remember - Keep Math simple. Don't complicate it! Plugging In can be a really helpful tool in some of the superficially complicated questions.
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