Proof of conjecture:if n is not prime ,then 2^n-1 is not prime

Proof of conjecture:if n is not prime ,then 2^n-1 is not prime

For n>1,and n is not prime, Prove that (2^n)-1 is not prime.

Assume n=ab ,where a < n ,b < n

Let x=(2^b)-1

Let y =1+(2^b)+(2^2b)+...+(2^(a-1)b

xy=( (2^b)-1)(1+(2^b)+(2^2b)+...+(2^(a-1)b) )

=(2^b+2^2b+2^3b+...+2^ab) -(1+(2^b)+(2^2b)+...+(2^(a-1)b)

xy=(2^ab)-1 = (2^n)-1

Since b< n,

(2^b)-1 < (2^n)-1

ie, x < (2^n)-1 //x is an integer less than (2^n)-1

Since ab=n and a

b>1

Since x=(2^b)-1 and b>1, x > (2^1)-1

ie , x>1 //x is greater than 1

xy=n

y1

y<(2^n)-1 //y is an integer less than (2^n)-1

Given y=1+(2^b)+(2^2b)+...+(2^(a-1)b

So, y>1 //y is greater than 1

Since,x is an integer less than (2^n)-1

and y is an integer less than (2^n)-1

and x is greater than 1

and y is greater than 1

and xy=(2^n)-1

(2^n)-1 can be written as the product of x and y which are greater than 1 and less than (2^n)-1

So,(2^n)-1 is not prime