Two natural numbers differ by 3 and their product is 504

Solution:

Let the required numbers be $x$ and $(x+3)$.

According to the question:

$x(x+3)=504$

$\Rightarrow x^{2}+3 x=504$

$\Rightarrow x^{2}+3 x-504=0$

$\Rightarrow x^{2}+(24-21) x-504=0$

$\Rightarrow x^{2}+24 x-21 x-504=0$

$\Rightarrow x(x+24)-21(x+24)=0$

$\Rightarrow(x+24)(x-21)=0$

$\Rightarrow x+24=0$ or $x-21=0$

$\Rightarrow x=-24$ or $x=21$

If $x=-24$, the numbers are $-24$ and $\{(-24+3)=-21\}$.

If $x=21$, the numbers are 21 and $\{(21+3)=24\}$.

Hence, the numbers are $(-24,-21)$ and $(21,24)$.