I've been working on this for a while now and I can't figure out how to prove it: $\left\lfloor \sqrt{2x} + \frac{1}{2}\right\rfloor = \left\lceil \frac{\sqrt{1+8x}-1}{2}\right\rceil.$
Here $x$ is an integer.
I've been working on this for a while now and I can't figure out how to prove it: $\left\lfloor \sqrt{2x} + \frac{1}{2}\right\rfloor = \left\lceil \frac{\sqrt{1+8x}-1}{2}\right\rceil.$
Here $x$ is an integer.
The result holds for every positive integer $x$.
Back to the definitions. Call $A_x$ and $B_x$ the LHS and the RHS of the relation to prove. Then $A_x=n$ if and only if $n$ is an integer and $ n\le\sqrt{2x}+\textstyle{\frac12}
What is missing from the steps above and what will save the day in the end is the fact that $x$ is an integer.
To wit, assume that $B_x=n+1$. Then, $\frac12(\sqrt{1+8x}-1)>n$ hence $x>\frac12n(n+1)$. Since $x$ and $\frac12n(n+1)$ are both integers, $ x\ge\textstyle{\frac12}n(n+1)+1. $ The inequality $\sqrt{2x}+\frac12
Added in note It is an interesting exercise to spot the point where the proof above goes astray for $x=0$. (Hint: as is often the case, almost right from the start.)