The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number.

Solution:

Let the greater number be $x$ and the smaller number be $y$.

According to the question:

$x^{2}-y^{2}=45 \quad \ldots($ i)

$y^{2}=4 x \quad \ldots$ (ii)

From (i) and (ii), we get:

$x^{2}-4 x=45$

$\Rightarrow x^{2}-4 x-45=0$

$\Rightarrow x^{2}-(9-5) x-45=0$

$\Rightarrow x^{2}-9 x+5 x-45=0$

$\Rightarrow x(x-9)+5(x-9)=0$

$\Rightarrow(x-9)(x+5)=0$

$\Rightarrow x-9=0$ or $x+5=0$

$\Rightarrow x=9$ or $x=-5$

$\Rightarrow x=9 \quad(\because x$ is a natural number $)$

Putting the value of $x$ in equation (ii), we get:

$y^{2}=4 \times 9$

$\Rightarrow y^{2}=36$

$\Rightarrow y=6$

Hence, the two numbers are 9 and 6 .