Difference between Perfect Squares

since $a>b$ , $a=b+1$

$a^2-b^2=501$

substitute:

$(b+1)^2-b^2=501$

$b^2+2b+1-b^2=501$

$2b=500$

$b=250$

$a$ and $b$ are consecutive positive integers, with $a>b$ , so $a-b=1$ . Thus

$\begin{aligned} a^2 - b^2 &= 501\\ (a+b)(a-b) &= 501\\ \implies a+b &= 501\\ a-b &= 1\\ \implies 2b &= 500\\ b&= 250\\ \end{aligned}$

a^2-B^2=501,so a+B=501(proof at the end),or B+(B+1)=501(The problem mentioned the numbers were consecutive,so a=B+1).Solving,we get B=250(B=B+1=501,B+B=500,2B=500,B=250).PROOF(involves good understanding of multiplication):Let's say a=B+1.So,to get from B^2 to a^2,we add B(there are now B+1(or a)groups of B).Then we switch the terms and add a(there are now B+1 B+1s or a sets of a or a^2).As you see,we added an a and a B to go from b^2 to a^2.Therefore a^2-b^2=a+b.You can actually generalize my proof to any number of consecutive numbers.For that the formula is a^2-b^2(where a=b+n)=b+2(b+1+b+2+b+3....b+(n-1))+a.Try proving it yourself!