If a^2-b^2 is a prime number, then show that a^2-b^2=a+b.where a and b are natural number
Answers
Answered by
6
If a number is prime number then only factors of that number is 1 and the same number
a^2-b^2=(a+b)(a-b).
So the factors are a+b and a-b
One these will be 1. Since a and b are natural numbers only possible chance is
a-b=1.
So a^2-b^2=(a+b) x 1
a^2-b^2=(a+b)
a^2-b^2=(a+b)(a-b).
So the factors are a+b and a-b
One these will be 1. Since a and b are natural numbers only possible chance is
a-b=1.
So a^2-b^2=(a+b) x 1
a^2-b^2=(a+b)
Similar questions