Question

✔✔a and b are two natural numbers such that a^2-b^2 is a prime number.Then the value of a^2-b^2 is =>

A)a-b
B)a+b
C)ab
D)None of the above.

STEP BY STEP EXPLANATION PLEASE

Answer 1

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

Now it's prime no

So there should not be factor 2

if a and b both are even or a and b both are odd then it becomes even

So one should be odd and other should be even

As ab gives 2 as factor making it even

So eliminate C option

for it to be prime one factor should be 1

as multiples of 2 integers would make it non prime

So either a+ b= 1 or a-b = 1

It would make it's value ( a-b) or ( a+b)

Now we have to choose between A and B option

it can't be a+b = 1

as both are natural no's

so minimum value is 2

So a+b

B) is correct option

100% sure

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