A's and B's

${ a }^{ 2 }\quad -\quad { b }^{ 2 }\quad =\quad 13\\ so\\ (a\quad -\quad b)(a\quad +\quad b)\quad =\quad 13\\ now\quad we\quad make\quad factors\quad of\quad 13\\ (a\quad -\quad b)(a\quad +\quad b)\quad =\quad 1\times 13\\ We\quad can\quad assume\quad \\ a\quad -\quad b\quad =\quad 1\quad \\ a\quad +\quad b\quad =13\\ Adding\quad the\quad we\quad get\\ 2a\quad =\quad 14\\ a\quad =\quad 7\quad \leftarrow ANS$ Quite Easily done QUITE EASILY DONE

(a+b)(a-b) =13 13 is a product of two numbers 13 & 1 then (a-b)=1 so, (a+b) =13 (a+b)+(a-b)=14 2a=14 a=7

a^2 _ b^2 = 13, so (a - b)(a + b) = 13, a prime. Then a - b = 1, a + b + 13, adding, 2a = 14, a = 7.

It is a rule of math that if two consecutive numbers will be added, the answer will be equal to the difference between the squares of those two numbers!

$a^2-b^2=13$ $(a+b)(a-b)=13$ As $a+b$ is bigger than $a-b$ and $13$ is prime so $a+b=13$ and $a-b=1$ .Solving this system of equations we get $a=7,b=6$ so $a=\boxed{7}$