if a^2-b^2 is prime then prove that a^2-b^2=a+b if a and b are +ve integers.

arre yaar...a²-b²=(a+b)(a-b) so (a+b) and (a-b) are two factors.....for a prime number the factor shud be either itself or 1

we cannot have (a+b)=1 as a and b are positive integers...

so we can have a-b=1

multiplying both sides by (a+b) we have (a²-b²) =(a+b) 

hence the problem....

and btw vatika i guess u got a so called "AIR 21"...na.....so go ahead and tell us wat branch u got....cmon'... 

a^2-b^2 can be written as (a+b)(a-b)

since a^2-b^2 is prime ..either (a+b) or (a-b) will be equal to 1..that too must be noted that the other one should be prime as well ...(bcos a and b are +ve integers)

hence two pts are clear...

1. one among (a+b) or (a-b) is equal to 1.

2. the other is prime...

Now...a and b are +ve integers...only (a-b) can be equalled to 1 ..and not (a+b)..

This does it ...

a^2 -b^2 =(a-b)(a+b)

                =(1)(a+b)

                =(a+b)

nice question ...!!