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

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