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Cel mai inteligent răspuns!
2014-11-04T14:54:44+02:00
Prin reducere la absurd pesupunem ca (2n+3 , 5n+7)≠1
fie d= (2n +3, 5n+7)
d/2n+3 | ·5⇒ d/ 10n+ 15 
d/ 5n+7 | ·2 ⇒ d/ 10n+ 14 ⇒ d/ 10n+15-(10n+14)
                                          d/1 ⇒d=1 ⇒ (2n+3, 5n+7)=1
2 5 2