Find two consecutive numbers whose squares differ by 31

Let x=smaller number (integer), then
x+1=larger number (integer)
Given
(x+1)^2-x^2=31
expand and simplify
x^2+2x+1-x^2=31
2x+1=31
x=15 (smaller number)
x+1=(larger number)

Check:
16^2-15^2=256-225=31  ok

the consecutive numbers are 15 and 16.

Let your two numbers be "a" and "b", where b=a+1.
You want
  b^2 - a^2 = 31
  (b + a)(b -a) = 31
  (a+1 + a)(1) = 31
  a = 15

The numbers are 15 and 16.

The correct answer is 7