Solve the following 7^x ≡13(mod 18) Pl give complete process.


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Question Number 13744 by AH Soomro last updated on 23/May/17

$S o l v e t h e f o l l o w i n g$ $7^{x} \equiv 13 (m o d 18)$ $P l g i v e c o m p l e t e p r o c e s s .$
Commented by prakash jain last updated on 23/May/17
$7≡7 (mod 18) 7^2 =49≡13 (mod 18) 7^3 =343≡1 (mod 18) 7^(3n) ≡1 (mod 18) 7^(3n+1) ≡7(mod 18) 7^(3n+2) ≡13 (mod 18) x=3n+2, n∈{0,1,2...}$
$7 \equiv 7 (\mod 18)$ $7^{2} = 49 \equiv 13 (\mod 18)$ $7^{3} = 343 \equiv 1 (\mod 18)$ $7^{3 n} \equiv 1 (\mod 18)$ $7^{3 n + 1} \equiv 7 (\mod 18)$ $7^{3 n + 2} \equiv 13 (\mod 18)$ $x = 3 n + 2, n \in {0, 1, 2 . . .}$
Commented by mrW1 last updated on 23/May/17

$\oint a n t a s t i C!$
Commented by RasheedSindhi last updated on 24/May/17

$We can come directly to$ $third step, because {Euler}^{'} s$ $theorem tells us that$ $If (a, m) = 1, then$ $a^{\emptyset (m)} \equiv 1 (\mod m)$ $[where \emptyset is Euler function$ $and \emptyset (m) is meant number of$ $coprimes of m which are not$ $greater than m .]$ $Hence 7^{3} \equiv 1 (\mod 18)$ $∵ \emptyset (7) = 3 and (7, 18) = 1$
Commented by AH Soomro last updated on 24/May/17

$Thanks to all .$
	Answered by mrW1 last updated on 23/May/17

	$7^{x} = 18 n + 13 = 7 \times 2 (n + 1) + 4 n - 1$ $4 n - 1 = 7 m$ $n = \frac{7 m + 1}{4}$ $\Rightarrow 7^{x} = 18 n + 13 = \frac{18 \times (7 m + 1) + 52}{4} = \frac{7 (9 m + 5)}{2}$ $\frac{9 m + 5}{2} = 7^{i}$ $m = \frac{2 \times 7^{i} - 5}{9}$ $f o r m t o b e i n t e g e r \Rightarrow i = 1, 4, 7 . . . o r 3 k - 2 w i t h k \in N^{+}$ $i . e . m = \frac{2 \times 7^{3 k - 2} - 5}{9} (* s e e c o m m e n t)$ $\Rightarrow 7^{x} = \frac{7 (18 m + 5)}{2} = 7 \times 7^{3 k - 2} = 7^{3 k - 1}$ $\Rightarrow S o l u t i o n i s x = 3 k - 1, k \in N^{+}$ $i . e . x = 2, 5, 8, 11 . . .$
	Commented by mrW1 last updated on 23/May/17

	$H e r e t h e p r o o f t h a t$ $m = \frac{2 \times 7^{3 k - 2} - 5}{9} = i n t e g e r$ $o r$ $2 \times 7^{3 k - 2} - 5 i s d i v i s i b l e b y 9 .$ $(1) c h e c k f o r k = 1$ $2 \times 7^{3 \times 1 - 2} - 5 = 2 \times 7 - 5 = 9 \equiv 0 (m o d 9)$ $\Rightarrow t r u e$ $(2) s u p p o s e d {i t}^{'} s t r u e f o r k$ $i . e . 2 \times 7^{3 k - 2} - 5 \equiv 0 (m o d 9)$ $f o r k + 1 w e h a v e :$ $2 \times 7^{3 (k + 1) - 2} - 5$ $= 2 \times 7^{3} \times 7^{3 k - 2} - 7^{3} \times 5 + 7^{3} \times 5 - 5$ $= 7^{3} \times (2 \times 7^{3 k - 2} - 5) + (7^{3} - 1) \times 5$ $= 7^{3} \times (2 \times 7^{3 k - 2} - 5) + 38 \times 9 \times 5 \equiv 0 (m o d 9)$ $\Rightarrow {i t}^{'} s t r u e f o r k + 1$ $p r o v e d!$
	Commented by AH Soomro last updated on 24/May/17

	$T h a n k s .$
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