How can one show the inequality Let $a,b,n$ be natural numbers (in $\mathbb{N}^*$) such that
$a>b$ and $n^2+1=ab$
How can one show that $a-b\geq\sqrt{4n-3}$, and for what values of $n$ equality holds?
I tried this:
We suppose that $a-b\geq\sqrt{4n-3}$ so
$$a^2-2ab+b^2\geq 4n-3\,.$$
And we have $ab=n^2+1$.
So $a^2-2(n^2+1)+b^2-4n+3\geq 0$.
So $a^2-2n^2+b^2-4n+1\geq 0$.
I'm stuck here! How do I proceed?
 A: Let $x:=a-n$ and $y:=b+x-n$ (that is, $a=n+x$ and $b=n-x+y$).  It can be easily seen that $x$ and $y$ are positive integers.  We want to find $a-b=2x-y$.  Note that $n^2+1=ab$ implies that $x^2-xy=ny-1$, or equivalently, $(2x-y)^2=y^2+4(ny-1)$.  As $y\geq 1$, we have $$(a-b)^2=(2x-y)^2\geq 1+4(n-1)=4n-3\,$$  Therefore, $a-b\geq \sqrt{4n-3}$.  The equality holds iff $y=1$, which leads to $$(n,a,b)=\left(x^2-x+1,x^2+1,x^2-2x+2\right)$$ for a positive integer $x$.
A: Substitute n with $\sqrt{(ab-1)}$
Then it should become easy.  
$
a-b \ge \sqrt{4\sqrt{(ab-1)}-3}
$
Now, keep raising to the power of 2 and see what happens.  
If in the last inequality you denote $k = a-b$ and you work with $b$ and $k$ only,
then it might become even easier. Just try, some of these routes should work.   
OK, if we implement the last idea, and we let $k=1$ and let $b$ go to infinity,
we can see that this inequality does not hold not true in general.
So I think the problem statement is wrong.    
A: Im just sharing an idea see $a>b$ thus $a-b>0$ now $0$ is less than all natural numbers now $\sqrt{4n-3}>=1$ as n belongs to natural numbers so least value it can have is $1$ . least value $a-b$  is also $1$ as difference between any two consecutive  numbers is smallest so we can say that $a-b>=1$ thus we can say that $a-b>=\sqrt{4n-3}$ hope you get what my basic idea is
