if $n^2Question:

There are different a number of natural numbers $a,b,c,d$,there such 
  $$n^2<a<b<c<d<(n+1)^2$$
  show that:

$$ad\neq bc$$
This problem is from this PDF (page 9,and the second problem): see 
link
My idea: Assume that there exist natural numbers $a,b,c,d$ such 
$$ad=bc$$
then I can't  have contradiction
Thank you

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.
