Given k ∈ N. 1) justify these relations: 3^(2k) +1≡2[8] and 3^(2k+1) +1≡4[8]. 2) Given (E): 2^n −3^m =1. n and m are unknowed. • Show that if m is even , (E) does not have solution. ■ Deduct from the first question 1) that the couple (2;1) is the only solution of (E).


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Question Number 119580 by mathocean1 last updated on 25/Oct/20

$Given k \in N .$ $1) justify these relations :$ $3^{2 k} + 1 \equiv 2 [8] and 3^{2 k + 1} + 1 \equiv 4 [8] .$ $2) Given (E) : 2^{n} - 3^{m} = 1 . n and m are unknowed .$ $∙ Show that if m is even, (E) does not have$ $solution .$ $◼ Deduct from the first question 1) that the$ $couple (2; 1) is the only solution of (E) .$
	Answered by mindispower last updated on 25/Oct/20

	$3^{2 k + 1} + 1 \equiv 4 [8] ?$
	Commented by mathocean1 last updated on 25/Oct/20

	$yes sir; or there is an error ?$
	Answered by mindispower last updated on 25/Oct/20
	$3^(2k) +1=9^k +1≡1^k +1=2[8] 3^(2k+1) +1=9^k .3+1≡3.(1)^k +1≡4]8] (E)⇔2^n −(3^m +1)=0 m=2k ⇒2^n =(1+3^(2k) )≡2[8] withe 1⇒n=1 but n=1 (E)⇒2−9^k =1⇒k=0 (E) has solution (n,m)=(1,0) m=2k+1 ⇒2^n =(1+3^(2k+1) )≡4(8)...withe Quation 1 ⇒n=2, 4−3.9^k =1⇒k=0 m=2.0+1=1,n=2 solution are(n,m)∈{(1,0);(2,1)}$
	$3^{2 k} + 1 = 9^{k} + 1 \equiv 1^{k} + 1 = 2 [8]$ $3^{2 k + 1} + 1 = 9^{k} . 3 + 1 \equiv 3 . {(1)}^{k} + 1 \equiv 4] 8]$ $(E) \Leftrightarrow 2^{n} - (3^{m} + 1) = 0$ $m = 2 k$ $\Rightarrow 2^{n} = (1 + 3^{2 k}) \equiv 2 [8]$ $w i t h e 1 \Rightarrow n = 1$ $b u t n = 1 (E) \Rightarrow 2 - 9^{k} = 1 \Rightarrow k = 0$ $(E) h a s s o l u t i o n (n, m) = (1, 0)$ $m = 2 k + 1$ $\Rightarrow 2^{n} = (1 + 3^{2 k + 1}) \equiv 4 (8) . . . w i t h e Q u a t i o n 1$ $\Rightarrow n = 2,$ $4 - 3 . 9^{k} = 1 \Rightarrow k = 0$ $m = 2.0 + 1 = 1, n = 2$ $s o l u t i o n a r e (n, m) \in {(1, 0); (2, 1)}$
	Answered by 1549442205PVT last updated on 25/Oct/20

	$1 a - We have 3^{2 k} + 1 = 9^{k} + 1 = {(8 + 1)}^{k} + 1$ $= \underset{m = 0}{\overset{k}{Σ}} C_{k}^{m} 8^{k - m} = (8^{k} + k . 8^{k - 1} + \frac{k (k -)}{2} 8^{k - 2}$ $+ . . . + 8 + 1) + 1 \equiv 2 (\mod 8) (q . e . d)$ $b - By above proof we have 3^{2 k} + 1 = 8 q + 2$ $\Rightarrow 3^{2 k + 1} + 1 = 3 . (3^{2 k} + 1) - 2 = 3 . (8 q + 2)$ $- 2 = 3 . 8 q + 4 \equiv 4 (\mod 8) (q . e . d)$ $2) a - Consider the equation 2^{n} - 3^{m} = 1$ $If m is even then 2^{n} = 3^{2 k} + 1 ()$ $(by above proof) . For n = 0, 1, 2 we see$ $the equation () has no roots$ $For n ⩾ 3 L . H . S () is divisible by 8$ $while R . H . S {isn}^{'} t divisible 8 since$ $3^{2 k} + 1 \equiv 2 (\mod 8) (by above proof) .$ $Hence this case the equation has no roots$ $(q . e . d)$ $b - We see that (by directly checking)$ $the (n, m) = (2, 1) satisfy the equation$ $2^{n} - 3^{m} = 1 (1) . We prove that is unique$ $solution of given equation . Indeed,$ $When m is even the equation has no$ $roots (above proof) . We now consider$ $the case m is odd . \Rightarrow m = 2 k + 1 . Then$ $(1) \Leftrightarrow 2^{n} = 3^{2 k + 1} + 1 ( )$ $For n = 0, 1 It is clear that the equation$ $( ) has no roots$ $For n ⩾ 2 L . H . S ( ) is divisible by 4$ $while R . H . S ( *) {isn}^{'} t divisible (by$ $above proof 1 b) . Hence this case the$ $given has no roots which means the$ $(n, m) = (2, 1) is unique solution of the$ $given equation (q . e . d)$
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