Question

The sum of the squares of two positive integer is 208. If the square of larger number is 18 times the smaller number, find the numbers.

Answer 1

Let's assume the two parts to be $x$ and $y$, $y$ being the larger of the two numbers.
Then, from the question
$x^2+y^2=208$ .......(i) and
$y^2=18x$ .......(ii)
From (i), we get $y^2=208-x^2$
Now, putting this in (ii), we have
$208-x^2=18x$
$x^2+18x-208=0$
$x^2+26x-8x-208=0$
$x(x+26)-8(x+26)=0$
As $x$ can't be a negative integer, so $x=8$ is valid solution.
Using $x=8$ in (ii), we get $y^2=18\times 8=144$
Thus, $y=12$ only as $y$ is also a positive integer
Therefore, the two numbers are 8 and 12.