Question

For a and b are positive integer, a > b, and b > 2. Can the value of :
(2^a + 1)/(2^b - 1) be an integer ? Give your reason ....

Answer 1

Let a = qb+r
such that r < b

Answer 2

I was take simpler, a = b + k
with k is positive integer. Is it same with yours @ganeshie8 ?

Answer 3

I want r to be less than b

Answer 4

a= b+k does not guarantee that k is less than b right?

Answer 5

2^a+1 = 2^(qb+r) + 1

= 2^r*2^(qb) + 1

= 2^r*(2^b)^q + 1

= 2^r*( `2^b-1` +1)^q + 1

Answer 6

But for b + k gives guarantee more than b ? Ok.. i will follow your idea :)

Answer 7

Next we use this:
(x+1)^n = 1 + nx + (x^2 and greater terms)

Answer 8

I'm on mobile, facing issues with latex

Answer 9

2^a+1 = 2^(qb+r) + 1

= 2^r*2^(qb) + 1

= 2^r*(2^b)^q + 1

= 2^r*( 2^b-1 +1)^q + 1

= 2^r(1+q(2^b-1)) + 1

= q(2^b-1) + 2^r+1

Answer 10

We can ignore the first term q(2^b-1) as it is divisible by 2^b-1

Answer 11

The rest 2^r+1 is clearly less than 2^b-1 whenever r < b

Answer 12

Yeah... i caught your explain. So, the answer is impossible be an integer right ?

Answer 13

We have just showed that 2^a +1 leaves a nonzero remainainder 2^r+1 when divided by 2^b-1

So the given expression can never be an integer

Answer 14

Yup. Thank you so much @ganeshie8

Answer 15

Np :)

Answer 16

2^a+1 = 2^(qb+r) + 1

= 2^r*2^(qb) + 1

= 2^r*( `2^b` )^q + 1

= 2^r*( `2^b-1 +1`)^q + 1
i don't understand the last line.. why did you replace 2^b with 2^b-1 `+1`