Question

Find two consecutive positive integers, the sum of whose squares is 25.

Answer 1

$$\begin{gathered} \text{Let the two consecutive positive integers be}x\mathrm{and}(x+1). \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}: \\ {x}^2+{(x+1)}^2=25 \\ \text{⇒}{x}^2+x^2+2x+1-25=0 \\ \text{⇒ 2}{x}^2+2x-24=0 \\ \text{⇒}{x}^2+x-12=0 \\ \text{⇒}{x}^2+(4-3)x-12=0 \\ \text{⇒}{x}^2+4x-3x-12=0 \\ \text{⇒}x(x+4)-3(x+4)=0 \\ \Rightarrow (x+4)(x-3)=0 \\ \therefore x=-4,x=3 \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{the}\mathrm{integers}\mathrm{are}\mathrm{positive};\mathrm{therefore}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{will}\mathrm{be}3. \\ \mathrm{Thus}\text{, the two positive integers are}3\text{and {(3 + 1) = 4}.} \end{gathered}$$