I think mreditor was referring to the solutions to this question
<a href="http://www.codecogs.com/eqnedit.php?latex=$&space;Show&space;that&space;if&space;\sqrt{n+\frac{1}{4}}>&space;k&space;-&space;\frac{1}{2}$&space;\&space;$then&space;the&space;integers&space;n&space;and&space;k&space;satisfy&space;\sqrt{n}>&space;k&space;-\frac{1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?$&space;Show&space;that&space;if&space;\sqrt{n+\frac{1}{4}}>&space;k&space;-&space;\frac{1}{2}$&space;\&space;$then&space;the&space;integers&space;n&space;and&space;k&space;satisfy&space;\sqrt{n}>&space;k&space;-\frac{1}{2}" title="$ Show that if \sqrt{n+\frac{1}{4}}> k - \frac{1}{2}$ \ $then the integers n and k satisfy \sqrt{n}> k -\frac{1}{2}" /></a>
When you square both sides you get
<a href="http://www.codecogs.com/eqnedit.php?latex=n>k(k-1)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?n>k(k-1)" title="n>k(k-1)" /></a>
Solutions add 1 to the RHS without explaining why (the dodgy part I assume). I was told by friends and a teacher that difference in the two integers in an equality must be greater than 1 or something, hence adding one to the RHS doesn't change the inequality. LHS still > RHS. So that means 1/4 can be added and won't change the inequality.
<a href="http://www.codecogs.com/eqnedit.php?latex=n>&space;k(k-1)+1&space;\\&space;\therefore&space;n>&space;k(k-1)+\frac{1}{4}\\&space;n>&space;(k-\frac{1}{2})^{2}&space;\\\\&space;\therefore&space;\sqrt{n}>(k-\frac{1}{2})" target="_blank"><img src="http://latex.codecogs.com/gif.latex?n>&space;k(k-1)+1&space;\\&space;\therefore&space;n>&space;k(k-1)+\frac{1}{4}\\&space;n>&space;(k-\frac{1}{2})^{2}&space;\\\\&space;\therefore&space;\sqrt{n}>(k-\frac{1}{2})" title="n> k(k-1)+1 \\ \therefore n> k(k-1)+\frac{1}{4}\\ n> (k-\frac{1}{2})^{2} \\\\ \therefore \sqrt{n}>(k-\frac{1}{2})" /></a>
