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

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 \dots

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 \dots .

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 \dots . .

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