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

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^{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) \dots 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}

+ \dots + 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)