i still search about a general and complete solution about this determine x in N where 7 divise 2^x +3^x note = it is just an exercise in secondary so dont go away... maybe we must use separation of cases methode....


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 21990 by hi147 last updated on 08/Oct/17

$i s t i l l s e a r c h a b o u t a g e n e r a l a n d$ $c o m p l e t e s o l u t i o n a b o u t t h i s$ $d e t e r m i n e x i n N w h e r e 7 d i v i s e 2^{x} + 3^{x}$ $n o t e = i t i s j u s t a n e x e r c i s e i n s e c o n d a r y$ $s o d o n t g o a w a y . . .$ $m a y b e w e m u s t u s e s e p a r a t i o n o f c a s e s$ $m e t h o d e . . . .$
Commented by Tinkutara last updated on 11/Oct/17

$I got x \equiv 3 (\mod 6) . Is this true ?$
	Answered by Tinkutara last updated on 11/Oct/17

	$B y {F e r m a t}^{'} s l i t t l e t h e o r e m,$ $2^{6} \equiv 1 (m o d 7)$ $\Rightarrow 2^{6 k + r} \equiv 2^{r} (m o d 7)$ $S i m i l a r l y, 3^{6 k + r} \equiv 3^{r} (m o d 7)$ $\Rightarrow 2^{6 k + r} + 3^{6 k + r} \equiv 2^{r} + 3^{r} (m o d 7)$ $F o r k = 0 a n d t r y i n g f o r 0 < r < 6,$ $2^{1} + 3^{1} \equiv 5 (m o d 7)$ $2^{2} + 3^{2} \equiv 6 (m o d 7)$ $2^{3} + 3^{3} \equiv 0 (m o d 7)$ $2^{4} + 3^{4} \equiv 6 (m o d 7)$ $2^{1} + 3^{1} \equiv 2 (m o d 7)$ $H e n c e x \equiv 3 (m o d 6) .$
	Commented by hi147 last updated on 16/Oct/17

	$t h a t s g o o d b u t h o w c a n w e p r o v e t h a t$ $i t i s t h e o n l y s o l u t i o n . i m e a n 6 k + 3 .$ $a n d w e h a v e a n o t h e r p r o b l e m = i t i s$ $j u s t a n e x e r c i s e i n s e c o n d a r y$ $a n d w e d o n t h a v e f e r m a t s l i t t e l t h e o r e m$ $i n t h i s l i v e l . s o . . . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum