Long time ago for an assignment, I submitted an alternative proof of $\sqrt 2$ not being rational along the following lines:
suppose $\frac{a^2}{b^2}=2$ then we must also have $a=b+n$ for $a,b,n$ positive integers.
$(b+n)^2=2b^2$
$b^2+n^2+2bn=2b^2$
$n^2+2bn-b^2=0$ At this point I managed to somehow show that no integer $n$ can satisfy this ( not sure if I used quadratic equation or not), in place of 2, substituting $k$ in the first equation lead to $n$ not being an integer whenever $k$ was not a square integer. without having to change anything in the proof.
I can not reconstruct the missing steps, what possible continuation in reasoning can work for this case?