Solve and explain the following mathematical problem.

hello
Solve and explain the following mathematical problem.

a2-b2 is prime number then a2-b2 is
a+b
a-b
a+b(a-b)
none of these

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Friend,
- The problem that you have raised here belongs to simple mathematical formula.
� (a2 -b2) = (a+b)(a-b)
� This formula has been of much use in factorization in arithmetic.
� However,if the result is always a prime number,then you you will obtain the result only when two numbers are consecutive.
� In this case,the possible result (a+b) will also hold true.
- Hence,clearly,you can go for the 'Third option' or the 'first option'.

- For convenience,you can try inserting the values , a=3 & b=2
- We have (a2- b2)=5
- Clearly,both (a+b) and (a+b)(a-b) satisfy the result.

- Another such result is possible when a=2 and b=2.

- For further relevance,you must try putting in different values of numbers,so as to obtain prime number as a result.

ALL THE BEST.
aman//

Hello,
If a^2-b^2 is prime number then a^2-b^2=(a+b)(a-b).
1) A^2-b^2=(a+b)(a-b)
Take R.H.S
(a+b)(a-b) (multiple a and b with a-b)
= a^2-ab+ab-b^2 (+ab-ab=0 both ill cancel)
= a^2 + 0- b^2 (finally)
= a^2 - b^2
=L.H.S
Therefore R.H.S is equal to L.H.S
2)a^2-b ^2=(a+b)
a^2-b^2*= (a+b) * 1 (finally)
a^2- b^2*= a+b

for example- a=1& b=2
then (a^2- b^2)=3
Both (a+b) and (a+b)(a-b) satisfy the result.

a2-b2 = (a+b) (a-b)


a2-b2 is prime number one of the two factors has to be 1. Since, a and b are natural numbers, thus a+b can never be 1. So, a-b = 1.


a2- b2 = (a+b) * 1


= a2- b2 = a+b