How to prove the equality of the floors of two expressions?

Question

I was doing this problem from Olimpiada de Mayo (Argentina) problem when I wonder if $$ \bigg\lfloor\frac{-1+\sqrt{1+8n}}{2}\bigg\rfloor = \bigg\lfloor\frac{-1+\sqrt{9+8n}}{2}\bigg\rfloor $$ if $n\geq 3$ and $9+8n$ isn't a perfect square.

Answer 1

There is no odd square strictly between $1+8n$ and $9+8n$ in any case. When there is also no even square between the two numbers, (and the larger one not a square) the two square roots have the same integer part.

When $$ 1+8n < 4 m^2 < 9 + 8n $$ we find $$ 1+8n = 2m - \delta \; , \; \; \; 9+8n = 2m + \epsilon \; ,$$ with $$ 0 < \delta , \epsilon < 1 $$

This is not as bad as it looks, because of the $-1$ additions. $$ \left\lfloor \frac{2m-1 \pm \delta}{2}\right\rfloor = \left\lfloor \frac{2m-2}{2}\right\rfloor = m-1 $$

Example, with $n=4$ we get $33 < 36 < 41,$ $\sqrt{33} \approx 5.744 \approx 6 - 0.255 \; , \; \; \; $ $\sqrt{41} \approx 6.403 \approx 6 + 0.403 \; . \; \; \; $ $$ $$ Then $\sqrt{33} -1\approx 4.744 \approx 5 - 0.255 \; , \; \; \; $ $\sqrt{41}-1 \approx 5.403 \approx 5 + 0.403 \; . \; \; \; $ $$ $$ Then $ \frac{\sqrt{33} -1}{2} \approx \frac{5 - 0.255}{2} \approx 2.5 - 0.127 \; , \; \; \; $ $\frac{\sqrt{41}-1}{2} \approx \frac{ 5 + 0.403}{2} \approx 2.5 + 0.2015 \; . \; \; \; $

and both have $2$ as integer part.