The sum of the squares of two consecutive positive odd integers is 2050. What is the smaller integer?

Correct answer: 31

Let the smaller integer be \(x\). The next consecutive odd integer is therefore \(x+2\). Thus, \(x^2+(x+2)^2=2050\), which gives \(2x^2+4x-2046=0\) or \(x^2+2x-1023=0\). Factoring, \((x-31)(x+33)=0\), so \(x=31\) or \(x=-33\). Since the integers are positive, \(x=31\). Therefore, the correct answer is 31. Exam tip: consecutive odd integers differ by 2, not by 1.