Since they are consecutive, meaning that one is another added to a unit.
\(a=b+1\)
Where: a>b
Now, we must mathematizing the problem:
\(a^2-b^2=a+b \\ (a+b)(a-b)=(a+b) \\ a-b=1 \\ \boxed {a=b+1}\)
Note that, the result was expected because we made "a^2-b^2" in order to give a positive number, since "a" and "b" are positive, then the sum of both should also give a positive number.
If you notice any mistake in English, please let me know, because I'm not native.