Question

Find two consecutive odd positive integers, sum of whose squares is 970.

Answer 1

Let one of the number be x then other number is x + 2.

Then according to question,

\[x^2 + \left( x + 2 \right)^2 = 970\]

\[ \Rightarrow x^2 + x^2 + 4x + 4 = 970\]

\[ \Rightarrow 2 x^2 + 4x - 966 = 0\]

\[ \Rightarrow x^2 + 2x - 483 = 0\]

\[ \Rightarrow x^2 + 23x - 21x - 483 = 0\]

\[ \Rightarrow x(x + 23) - 21(x + 23) = 0\]

\[ \Rightarrow (x - 21)(x + 23) = 0\]

\[ \Rightarrow x - 21 = 0 \text { or } x + 23 = 0\]

\[ \Rightarrow x = 21 \text { or } x = - 23\]

Since, x being an odd positive integer,

Therefore, x = 21.

Then another number will be \[x + 2 = 21 + 2 = 23\]

Thus, the two consecutive odd positive integers are 21 and 23.