The sum of the squares of two consecutive odd numbers is 394. Find the numbers.

Let the two odd numbers be 2𝑥 + 1 and 2𝑥 + 3 The sum of the squares of two consecutive odd numbers is 394.

Therefore, $(2x+1{)}^2+(2x+3{)}^2=394$

$\begin{array}{r}\Rightarrow 4x^2+4x+1+4x^2+12x+9=394\\ \Rightarrow 8\left(x^2+2x-48\right)=0\\ \Rightarrow \left(x^2+2x-48\right)=0\end{array}$

Factorising by splitting the middle term, we get

$\Rightarrow \left(x^2+8x-6x-48\right)=0$

Taking terms common, 

⇒ 𝑥(𝑥 + 8) − 6(𝑥 + 8) = 0 

⇒ (𝑥 + 8)(𝑥 − 6) = 0 

⇒ 𝑥 + 8 = 0 or, 𝑥 − 6 = 0 

⇒ 𝑥 =− 8 or 6 

We can reject 𝑥 =− 8 as 2𝑥 + 1 and 2𝑥 + 3 will become negative.

Therefore, first odd number = 2𝑥 + 1 = 2(6) + 1 = 13. 

The other odd number = 2𝑥 + 3 = 2(6) + 3 = 15 

Hence, the two numbers are 13 and 15.