Bunuel wrote:

If x ≠ 0, does x have an odd number of factors?

(1) √x is an integer.

(2) x^2 is an integer.

Beautiful problem!

\(?\,\,\,\,:\,\,\,\,\,\# \,\,\left( {{\text{positive}}} \right)\,\,{\text{factors}}\,\,{\text{odd}}\,\,{\text{?}}\)

Important: it is not known (pre-statements) whether x is an integer (this is part of the problem)!

If x is not an integer, the answer (=focus) is <NO>, because non-integers don´t have factors (by definition, factors are related to integers only)!

Statement (2) will explore that. Let´s start with it!

\(\left( 2 \right)\,\,\,\left\{ \begin{gathered}

\,Take\,\,x = \sqrt 2 \,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\,\,\,\, \hfill \\

\,Take\,\,x = 1\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\

\end{gathered} \right.\)

\(\left( 1 \right)\,\,\sqrt x \,\,\,\operatorname{int} \,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{gathered}

x \geqslant 0\,\,\,\,{\text{implicitly}}\,\,\,\,\,\,\mathop \Rightarrow \limits^{x\,\, \ne \,\,0} \,\,\,\,\,x > 0 \hfill \\

x\,\, = {\left( {\sqrt x } \right)^2}\,\, = \,\,{\operatorname{int} ^{\,2}} = {\text{perfect}}\,\,{\text{square}} \hfill \\

\end{gathered} \right.\,\)

\(x\,\,{\text{perfect}}\,\,{\text{square}}\,\,\, \geqslant \,\,\,{\text{1}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\{ \begin{gathered}

\,\,x = 1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\, \hfill \\

\,\,x \geqslant 4\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{x}}\,{\text{ = }}\,{\text{prime}}{{\text{s}}^{\,{\text{even}}\,{\text{powers}}}}\,\,\, \hfill \\

\end{gathered} \right.\)

\({\text{x}}\,{\text{ = }}\,{\text{prime}}{{\text{s}}^{\,\boxed{{\text{even}}\,\,{\text{powers}}}}}\,\,\,\,\, \Rightarrow \,\,\,\,? = \left( {\boxed{{\text{even}}}\, + 1} \right)\,\, \cdot \left( {\boxed{{\text{even}}}\, + 1} \right) \cdot \ldots \cdot \left( {\boxed{{\text{even}}}\, + 1} \right) = {\text{odd}}\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\,\)

Important: the explanation above shows the reason why every nonzero perfect square has an odd number of positive factors.

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

_________________

Fabio Skilnik ::

GMATH method creator (Math for the GMAT)

Our high-level "quant" preparation starts here:

https://gmath.net