A number theory problem by Pramita Kastha

$a^2-b^2=23$

$\Rightarrow (a-b)(a+b)=23$

Because $23$ is a prime number, therefore $23=23\times1$

$(a-b)(a+b)=23\times1$

In natural numbers, addition of two numbers is always larger than subtraction:

$(a-b)=1$

$(a+b)=\boxed{23}$

You can stop here unless you want to find the value. If $a-b=1 \Rightarrow a=b+1$ , $a+b=23 \Rightarrow a=23-b$

$\Rightarrow b+1=23-b$

$2b=24 \Rightarrow b=\boxed{12} \Rightarrow a=\boxed{11}$

$a^{2}-b^{2} = 23$

$(a-b)(a+b) = 1 \times 23$

Clearly,

$(a-b) < (a+b)$

Hence,

$(a-b) = 1$

$(a+b) = 23$

23 can only be divided by 1 after solving equation. as we solve we would get 12 &11

Given that $a^{2} -b^{2} = 23$ and that Consecutive squares differ by odd integers , we are looking for $n$ such that $2 n+1=23$ . Thus, $n=11$ which means that $a=12$ and $b=11$ . Therefore, $a + b = 23$ .

The difference between any two squares n and (n+1) is (2n + 1).

Then, looking at 23 +/- 1, you have A, B (respectively) 23 + 1 = 24 24/2 = 12 A = 12

23 - 1 = 22 22/2 = 11 B = 11

A + B = 12 + 11 = 23

Interestingly, (a^2 - b^2) always = (a + b) as long as a = (b + 1).