Question

Find the smallest solution in positive integers of $x^2-7y^2-9=0$.

Answer 1

The given equation is $x^2-7y^2+9=0$

$\Rightarrow$ $x^2=7y^2+9$

The smallest possible positive integer for $y$ is $1$

For $y=1$

$\Rightarrow$ $x^2=7(1{)}^1+9$

$\Rightarrow$ $x^2=16$

$\therefore$ $x=4$

$\therefore$ The smallest solution in positive integers are $x=4$ and $y=1$