Proving that $\Bigl[\frac{-4+\sqrt{26+20y+4y^2}}{2},\frac{-3+\sqrt{23+20y+4y^2}}{2}\Bigl]$ contains no integer

Question

Let $y$ be a non-negative integer. How do I prove that the interval

$$\Biggl[\frac{-4+\sqrt{26+20y+4y^2}}{2},\frac{-3+\sqrt{23+20y+4y^2}}{2}\Biggl]$$

contains no integer? I've found out using divisibility reasoning that these bounds are both irrational, so any integer must lie in-between the bounds.

Answer 1

Hint: The interval can be written as $$[f(y),g(y)]= \Biggl[\frac{\sqrt{(2y+5)^2+1}}{2}-2, \frac{\sqrt{(2y+5)^2-2}}{2}-{3\over2}\Biggl] $$ It is not difficult to show that $$ f(y)=y+{1\over2}+\delta(y), \qquad g(y)=y+1-\varepsilon(y), $$ where both $\delta(y)$ and $\varepsilon(y)$ are positive, with $\delta(y)\to0$ and $\varepsilon(y)\to0$ as $y\to\infty$.

Answer 2

$$\Biggl[\frac{-4+\sqrt{26+20y+4y^2}}{2},\frac{-3+\sqrt{23+20y+4y^2}}{2}\Biggl]=\\\Biggl[\frac{-4+\sqrt{(2y+5)^2+1}}{2},\frac{-3+\sqrt{(2y+5)^2-2}}{2}\Biggl]$$ $y \in \left\{0,1,2,3,4,...\right\}$ you may know $\sqrt{x^2+\Delta x}\sim x+\dfrac{\Delta x}{2x}$ so $$\color{red} {\sqrt{(2y+5)^2+1}\sim 2y+5 +\dfrac{1}{4y+10}\\ \sqrt{(2y+5)^2-2}\sim 2y+5 +\dfrac{-2}{4y+10} }$$ $$\Biggl[\frac{-4+\sqrt{(2y+5)^2+1}}{2},\frac{-3+\sqrt{(2y+5)^2-2}}{2}\Biggl]=\\ \Biggl[\frac{-4+2y+5 +\dfrac{1}{4y+10}}{2},\frac{-3+2y+5 +\dfrac{-2}{4y+10}}{2}\Biggl]=\\ \Biggl[y+\dfrac12+\dfrac{1}{8y+20},y+1-\dfrac{2}{8y+20}\Biggl]\subseteq [y+0.5+0.05,y+1-0.1]$$ because min $y=0 \to \dfrac{1}{8y+20}\leq \frac{1}{20}$ so
$$\Biggl[y+\dfrac12+\dfrac{1}{8y+20},y+1-\dfrac{2}{8y+20}\Biggl]\subseteq \color{red} {[y+0.55,y+0.9]}\\ y<y+0.55<y+0.9<y+1$$ there is no integer between $y,y+1$