I have to show that there is a unique integer within the interval $[a,b]$ where

$a = -\frac{3}{2} + \sqrt{\frac{9}{4} + 2(n+1)}$


$b = -\frac{1}{2} + \sqrt{\frac{1}{4} + 2(n+1)}$

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?

  • $\begingroup$ It might be worth seeing when $a$ and $b$ are integers. It then becomes easy $\endgroup$
    – Henry
    Apr 28 '20 at 13:59
  • $\begingroup$ But that won't prove for the cases when a and b are not integers. I think we can break the problem into 2 parts.... $\endgroup$ Apr 28 '20 at 14:01
  • $\begingroup$ Wait, I think you went wrong somewhere, @Octavius. If you can prove 1=|b-a| for all a and b which are not integers, then there will definitely be an integer between a and b when and b are themselves not integers. $\endgroup$ Apr 28 '20 at 14:10
  • $\begingroup$ If the difference between any 2 real numbers a and b is greater than 1, then there will be at least one integer between [a, b]. $\endgroup$ Apr 28 '20 at 14:15
  • $\begingroup$ yes, but I proved that the difference (b-a) is smaller than 1, i.e. (b-a) < 1. $\endgroup$
    – Octavius
    Apr 28 '20 at 14:22

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.

  • $\begingroup$ An elegant solution, which I accept gratefully. Thank you. $\endgroup$
    – Octavius
    Apr 28 '20 at 18:29

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.


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