VJesus12 wrote:Is x^2 + 9 prime?
(1) x is odd
(2) 3 ≤ x ≤ 7
Source: Veritas Prep
$${x^2} + 9\,\,\,\mathop = \limits^? \,\,\,{\rm{prime}}$$
$$\left( 1 \right)\,\,\,x\,\,{\rm{odd}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^2} + 9\,\, \ge \,\,\,10\,\,\,{\rm{and}}\,\,{\rm{even}}\,\,\,\,\,\left( {{\rm{also}}\,\,{\rm{when}}\,\,x \le - 1} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\,\,$$
$$\left( 2 \right)\,\,\,3 \le x \le 7\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 3\,\,\,\, \Rightarrow \,\,\,{x^2} + 9 = 18\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = \,\,\sqrt {10} \,\,\,\left( {\sqrt 9 < \,\,\sqrt {10} < \sqrt {49} } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^2} + 9 = 19\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.