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
 A: 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.
A: 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.
A: Here is my attempt. It is imperfect but maybe it will lead to a solution. Define: 


*

*$A=a-n^2-n$

*$B=b-n^2-n$

*$C=c-n^2-n$

*$D=d-n^2-n$


Then the assumption becomes:
$-n<A<B<C<D<(n+1)$
And the contradiction assumption becomes:
$AD+(A+D)(n^2+n)=BC+(B+C)(n^2+n)$
Hence: $AD<BC$ and $(A+D)>(B+C)$.
They are all integers so: $(A+D)\geq(B+C+1)$ and $AD+n^2+n\leq BC$
From the assumption: $-n^2<AD<n^2$ and  $-n^2<BC<n^2$
Hence:  $(A+D)=(B+C+1)$ and $AD+n^2+n= BC$
Here I am stuck.
A: Here is a new video by Tim Gowers solving this problem in real time:
https://www.youtube.com/watch?v=NmEVwJ_lJ1A
