The sum of the squares of two consecutive odd positive integers is 1570. What is the smaller integer?

Correct answer: 27

Let the smaller integer be \(x\). The next consecutive odd integer is \(x+2\). Thus, \(x^2+(x+2)^2=1570\), which gives \(2x^2+4x-1566=0\) or \(x^2+2x-783=0\). Factoring, \((x-27)(x+29)=0\), so \(x=27\) or \(-29\). Since the integers are positive, the smaller one is 27. Check: \(27^2+29^2=729+841=1570\). Exam tip: consecutive odd integers always differ by 2.