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

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.
