help-x-N-determine-x-where-7-divise-2-x-3-x-

Question Number 21795 by hi147 last updated on 04/Oct/17

h e l p

x \in N

d e t e r m i n e x w h e r e 7 d i v i s e 2^{x} + 3^{x}

Commented by Rasheed.Sindhi last updated on 04/Oct/17

x = 3, 9

Commented by prakash jain last updated on 04/Oct/17

How did you get x = 3 ?

Commented by math solver last updated on 05/Oct/17

h o w d i d y o u g e t x = 9

Answered by mrW1 last updated on 04/Oct/17

x = 3^{n}, n \in N

prove :

for n = 1,

2^{3^{1}} + 3^{3^{1}} = 8 + 27 = 35 \mod 7 = 0

{it}^{'} s true for n = 1 .

{let}^{'} s assume {it}^{'} s true for n, i . e .

2^{3^{n}} + 3^{3^{n}} \mod 7 = 0

for n + 1 :

(2^{3^{n + 1}} + 3^{3^{n + 1}}) \mod 7

= (2^{3 \times 3^{n}} + 3^{3 \times 3^{n}}) \mod 7

= [{(2^{3^{n}})}^{3} + {(3^{3^{n}})}^{3}] \mod 7

= [(2^{3^{n}}) + (3^{3^{n}})] [{(2^{3^{n}})}^{2} - (2^{3^{n}}) (3^{3^{n}}) + {(3^{3^{n}})}^{2}] \mod 7

= [(2^{3^{n}}) + (3^{3^{n}})] \mod 7

= 0

\Rightarrow x = 3^{n} is a solution

Commented by hi147 last updated on 04/Oct/17

t h x

b u t h o w u t h i n k i n 3^{n} e x a c t l y ?

t h e p r o v e i s n o t c o m p l e t e \dots

i t s e e m s t h a t w e m u s t n o t u s e

c o n g r u e n c e m e t h o d e b e c a u s e t h i s

e x e r c i s e i s u n d e r t h e t i t l e (d i v i s i b i l i t y)

i n s e c o n d a r y b o o k . s o w e m u s t j u s t

u s e (n d i v i s e x \Leftrightarrow x = k n / k \in Z) \dots

i w a i t f o r a n o t h e r a n s w e r \dots

Commented by hi147 last updated on 04/Oct/17

t h e q u e s t i o n i s f r o m w h e r e c o m e 3^{n} ?

i m e a n b e f o r e u s i n g t h e d e m o n s t r a t i o n

w i t h c o n c u r r e n c e \dots . i f w e s a y f r o m

i m a g i n a t i o n . i t i s n o t a p r o v e . a n d a s

y o u s a y . m a y b e w e h a v e o t h e r x . s o

w e s t i l l n e e d a c o m p l e t s o l u t i o n \dots t h x

Commented by mrW1 last updated on 04/Oct/17

with x = 3^{n} we have proved that

7 divides 2^{x} + 3^{x} .

this prove is complete .

but indeed it is not proved that no

other solution than 3^{n} exists .

Commented by mrW1 last updated on 05/Oct/17

by try and error we get a solution x = 3 .

since a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2}) we

know that x = 3 \times 3 is then also a

solution, and x = 3 \times 3 \times 3 is also a

solution, etc . that means x = 3^{n} is

a solution .

Commented by mrW1 last updated on 05/Oct/17

generally 7 divides {(2^{3})}^{2 k + 1} + {(3^{3})}^{2 k + 1}

so x = 3 (2 k + 1) is a solution, k ⩾ 0 .

i . e . x = 3, 9, 15, 21, 27 \dots \dots

Commented by hi147 last updated on 06/Oct/17

g o o d . b u t h o w d i d u g e t {(2^{3})}^{2 k + 1} + {(3^{3})}^{2 k + 1}

a n d h o w w e c a n p r o v e t h a t 3 (2 k + 1) i s

t h e o n l y s o l u t i o n

t h x .

Commented by mrW1 last updated on 06/Oct/17

2^{3 (2 k + 1)} + 3^{3 (2 k + 1)}

= {(2^{3})}^{2 k + 1} + {(3^{3})}^{2 k + 1} = (2^{3} + 3^{3}) (\dots \dots .)

= 35 \times (\dots \dots)

= 7 \times 5 \times (\dots \dots)

\Rightarrow x = 3 (2 k + 1) is a solution .

I {don}^{'} t know if this is the only solution .

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