help x∈N determine x where 7 divise 2^x +3^x


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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

	$t h x$ $b u t h o w u t h i n k i n 3^{n} e x a c t l y ?$ $t h e p r o v e i s n o t c o m p l e t e . . .$ $i t s e e m s t h a t w e m u s t n o t u s e$ $c o n g r u e n c e m e t h o d e b e c a u s e t h i s$ $e x e r c i s e i s u n d e r t h e t i t l e (d i v i s i b i l i t y)$ $i n s e c o n d a r y b o o k . s o w e m u s t j u s t$ $u s e (n d i v i s e x \Leftrightarrow x = k n / k \in Z) . . .$ $i w a i t f o r a n o t h e r a n s w e r . . .$
	Commented by hi147 last updated on 04/Oct/17

	$t h e q u e s t i o n i s f r o m w h e r e c o m e 3^{n} ?$ $i m e a n b e f o r e u s i n g t h e d e m o n s t r a t i o n$ $w i t h c o n c u r r e n c e . . . . i f w e s a y f r o m$ $i m a g i n a t i o n . i t i s n o t a p r o v e . a n d a s$ $y o u s a y . m a y b e w e h a v e o t h e r x . s o$ $w e s t i l l n e e d a c o m p l e t s o l u t i o n . . . t h x$
	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

	$generally 7 divides {(2^{3})}^{2 k + 1} + {(3^{3})}^{2 k + 1}$ $so x = 3 (2 k + 1) is a solution, k ⩾ 0 .$ $i . e . x = 3, 9, 15, 21, 27 . . . . . .$
	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

	$2^{3 (2 k + 1)} + 3^{3 (2 k + 1)}$ $= {(2^{3})}^{2 k + 1} + {(3^{3})}^{2 k + 1} = (2^{3} + 3^{3}) (. . . . . . .)$ $= 35 \times (. . . . . .)$ $= 7 \times 5 \times (. . . . . .)$ $\Rightarrow x = 3 (2 k + 1) is a solution .$ $I {don}^{'} t know if this is the only solution .$
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