Question

Find two consecutive positive integers, the sum of whose squares is 365.

Answer 1

$$\begin{gathered} \text{Let the two positive integers be}x\mathrm{and}(x+1). \\ \text{According to the question:} \\ {x}^2+{(x+1)}^2=365 \\ \Rightarrow x^2+x^2+2x+1=365 \\ \Rightarrow 2x^2+2x-364=0 \\ \Rightarrow 2(x^2+x-182)=0 \\ \Rightarrow x^2+x-182=0 \\ \Rightarrow x^2+(14-13)x-182=0 \\ \Rightarrow x^2+14x-13x-182=0 \\ \Rightarrow x(x+14)-13(x+14)=0 \\ \Rightarrow (x+14)(x-13)=0 \\ \Rightarrow x+14=0\text{or}x-13=0 \\ \Rightarrow x=-14\text{or}x\text{= 13} \\ \Rightarrow x\text{= 13}(\because x\text{is a positive integer}) \\ \text{Hence, the numbers are 13 and (13 + 1 = 14).} \end{gathered}$$