Probus wrote:

If m is a positive integer, is m a perfect square.

(1) \(2\)\((m-1)^2\)\(-\)\((m-2)^2\)\(+\)\(2\)= \(m^2\)

(2) \((\sqrt{m}-1)(\sqrt{m}+1)\) = \(48\)

\(m \geqslant 1\,\,\,\operatorname{int}\)

\(m\,\,\,\mathop { = \,}\limits^? \,\,{\left( {\operatorname{int} } \right)^2}\)

(1) Expanding the LHS we get m^2, hence all values of m (integers or not, by the way) satisfy this statement. INSUFFICIENT

(Take m = 1 and m = 2 for one possible BIFURCATION.)

(2) \(LHS\,\, = \,\,{\left( {\sqrt m } \right)^2} - {\left( 1 \right)^2} = m - 1\) hence m-1 = 48 and m = 49. SUFFICIENT (<YES>)

Answer: B

The above follows the notations and rationale taught in the GMATH method.

Regards,

fskilnik.

_________________

Fabio Skilnik :: https://www.GMATH.net (Math for the GMAT)

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