# Find two natural numbers which differ by 3

Question:

Find two natural numbers which differ by 3 and whose squares have the sum 17.

Solution:

Let one natural number be $x$ and other $(x-3)$.

Then according to question

$(x)^{2}+(x-3)^{2}=117$

$x^{2}+x^{2}-6 x+9=117$

$2 x^{2}-6 x+9-117=0$

$2 x^{2}-6 x-108=0$

$2 x^{2}-6 x-108=0$

$2\left(x^{2}-3 x-54\right)=0$

$\left(x^{2}-3 x-54\right)=0$

$x^{2}-9 x+6 x-54=0$

$x(x-9)+6(x-9)=0$

$(x-9)(x+6)=0$

$(x-9)=0$

$x=9$

$(x+6)=0$

$x=-6$

Since, being a natural number, so x cannot be negative.

Therefore,

When $x=9$ then even integer

$x-3=9-3$

$=6$

Thus, two natural number be 9,6