I have to show that there is a unique integer within the interval $[a,b]$ where
$a = -\frac{3}{2} + \sqrt{\frac{9}{4} + 2(n+1)}$
and
$b = -\frac{1}{2} + \sqrt{\frac{1}{4} + 2(n+1)}$
as well as $n \in \mathbb{N}$.
Now the uniqueness is easy by showing that there can be one integer within this interval at maximum. This is done by calculating $(b-a)$ and finding that $(b-a)<1$.
But how do I prove that there indeed exists one?
On
I am down to something so I thought I should share it. It is given (or can be proved) b>a. Let a = m-h, b=m+k, where m is the integer unique integer lying in [a,b] and h and k are real positive numbers. We can (or have already above) prove that h and k are not integers that their sum is less than 1. First let n+1 be k. Reduce a and b to a form of k. Solving that, we get (2a+3)^2 = (2b+1)^2 + 8.
It reduces to (a+b+2)(a-b+1) = 2.
Putting in values for a and b, we get
(2m+h-k+2)(h+k+1) = 2.
This reduces to proving that m is indeed an integer. Hope this can help somehow.
Please tell if any mistakes are there.
Assume the contrary - $\nexists$ an integer lying in $[a,b]$. Then, $\exists\ m \in \mathbb{N}$ such that $[a,b] \subset [m,m+1]$.
So, \begin{align} m &< a \\ \implies m+\frac{3}{2} &< \sqrt{\frac{9}{4} + 2(n+1)} \\ \implies m^2 + 3m &< 2n+ 2 \end{align} and \begin{align} b &< m+1 \\ \implies \sqrt{\frac{1}{4} + 2(n+1)} &< m+\frac{3}{2} \\ \implies 2n +2 &< m^2 + 3m + 2 \\ \implies 2n &< m^2 + 3m\end{align}
Hence, we must have $m(m+3) = 2n+1$. But, $m(m+3)$ is an even number, while $2n+1$ is odd.
This is a contradiction, and so the result is proved.