if $n^2<a<b<c<d<(n+1)^2$ ,then prove $ad\neq bc$

This is a purely geometric inequality. I will prove that the only couple of points in $R=\{(x,y)\in\mathbb{N}^2: y\geq x, n^2\leq x,y\leq (n+1)^2\}$ collinear with the origin is $(n^2,n^2+n),(n^2+n,(n+1)^2)$.

Let $R_k$ be the subset of $R$ made of points for which $y-x=k>0$.

The arguments ($\arctan(y/x)$) of the elements of $R_k$ are clearly all distinct.

Now I claim that the argument of the rightmost element of $R_{k+1}$ is greater than the argument of the leftmost element of $R_k$. This is equivalent to: $$ \frac{n^2+k}{n^2} \leq \frac{(n+1)^2}{(n+1)^2-(k+1)},\tag{1}$$ or just to: $$ (n-k)^2 \geq 0,$$ with equality only when $k=n$. However, in order to have $ad-bc=0$, $(a,b)$ and $(c,d)$ must be two points of $R$ collinear with the origin. My previous argument now gives $b=c$, contradiction.

Claim: When $a<b<c<d$ and $ad=bc$ then there is a square $m^2$ with $a\leq m^2\leq d$.

Proof: Write $$a=r u,\quad b= r v,\qquad {\rm gcd}(u,v)=1\ ,$$ ensuring $v\geq u+1$. From $ad=bc$ it then follows that $c=su$, $d=sv$ with $s\in{\mathbb N}_{\geq1}$. Write $$r=t r',\quad s=t s',\qquad {\rm gcd}(r',s')=1\ .$$ Then $$a=t r' u,\quad b=t r' v,\quad c=t s' u,\quad d= ts' v\ ;$$ and as $b<c$ we conclude that $s'>r'$. It follows that $$d\geq t(r'+1)(u+1)\geq t r'u +2\sqrt{t}\sqrt{r' u}+1=\bigl(\sqrt{tr' u}+1\bigr)^2\ ;$$ whence $m:=\left\lceil\sqrt{t r'u}\ \right\rceil=\left\lceil\sqrt{a}\ \right\rceil$ does the job.