How many solutions!!!

Algebra Level 3

How many positive integers a , b a,b are there such that
2 a + 1 2^a+1 is divisible by 2 b − 1 2^b-1
Provided b ≥ 3 b \geq 3


The answer is 0.

Dinesh Chavan by Dinesh Chavan , 23, India

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1 solution

Xuming Liang
Mar 29, 2014

Since b = 1 b=1 always works, let b ≥ 2 b\ge 2 .

Consider the sum of the two numbers: 2 a + 1 + 2 b − 1 = 2 a + 2 b = 2 b ( 2 a − b + 1 2^a+1+2^b-1=2^a+2^b=2^{b}(2^{a-b}+1 Since 2 b − 1 2^b-1 divides the LHS and not 2 b 2^{b} , therefore 2 b − 1 ∣ 2 a − b + 1 2^b-1\mid 2^{a-b}+1 . We can continue this process until we get 2 b − 1 ∣ 2 a − b t + 1 = 2 x + 1 2^b-1\mid 2^{a-bt}+1=2^{x}+1 where a − b t = x a-bt=x and 0 < x < b 0<x<b (Note that if x = 0 x=0 then 2 b − 1 ∣ 2 2^b-1\mid 2 which only works when b = 1 b=1 . This means 2 x + 1 ≥ 2 b − 1 ⇒ 2 b − 1 + 2 ≥ 2 b ⇒ 2 b − 2 + 1 = 2 b − 1 2^{x}+1\ge 2^b-1\Rightarrow 2^{b-1}+2\ge 2^b\Rightarrow 2^{b-2}+1=2^{b-1} , so b = 2 b=2 is the only positive solution and since b ≥ 3 b\ge 3 therefore 0 \boxed {0} is the answer.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get 2 a + 2 b = 2 b t ( 2 x + 1 ) 2^a + 2^b = 2^{bt}(2^x + 1) ?

Isn't 2 a + 2 b = 2 b t + x + 2 b = 2 b t × 2 x + 2 b t 2^a + 2^b = 2^{bt + x} + 2^b = 2^{bt} \times 2^x + 2^bt ?

Siddhartha Srivastava - 7 years, 2 months ago

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I thought ahead of myself on that one, thanks for correcting it's now fixed!

Xuming Liang - 7 years, 2 months ago

@Xuming Liang Correct and nice solution

Dinesh Chavan - 7 years, 2 months ago

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