Question

Find two natural numbers which differ by 3 and the sum of whose squares is 117.

Answer 1

let the 2 numbers be $a,b$

Given that

$a-b=3$ $\left(1\right)$

$a^2+b^2=117$ $\left(2\right)$

Eq1: $a-b=3$

Squaring on both sides

$(a-b{)}^2=9$

$a^2+b^2-2ab=9$

substituting the value of $a^2+b^2$ from eq $\left(2\right)$ in eq $\left(1\right)$

$117-2ab=9$

$2ab=108$

$ab=54$ $\left(3\right)$

$b=\frac{54}{a}$

substituting this value of b in eq$\left(1\right)$, we get

$a-\frac{54}{a}$= $3$

$a^2-54=3a$

$a^2-3a-54$ = $0$

Solving this quadratic equation, we get

$a=9,-6$

but -6 is not the solution because it is given that they are natural numbers

So, $a=9$

Therefore,$b=9-3=6$

$b=6$